Exam 5: Sequences, Mathematical Induction, and Recursion

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A sequence b1,b2,b3,…b _ { 1 } , b _ { 2 } , b _ { 3 } , \ldots satisfies the recurrence relation bk=2bk−1+8bk−2b _ { k } = 2 b _ { k - 1 } + 8 b _ { k - 2 } with initial conditions b1=1b _ { 1 } = 1 and b2=0b _ { 2 } = 0 . Find an explicit formula for the sequence.

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The characteristic equation is t2−2t−8=0t ^ { 2 } - 2 t - 8 = 0 . Since t2−2t−8=(t−4)(t+2)t ^ { 2 } - 2 t - 8 = ( t - 4 ) ( t + 2 ) , there are two roots: t=4t = 4 and t=−2t = - 2 By the distinct roots theorem, there exist constants CC and DD such that
bn=C⋅4n+D⋅(−2)nb _ { n } = C \cdot 4 ^ { n } + D \cdot ( - 2 ) ^ { n } \quad for all integers n≥0.n \geq 0 .
Since b0=1b _ { 0 } = 1 and b1=0b _ { 1 } = 0 , then
{b0=C⋅40+D⋅(−2)0=C+D=1b1=C⋅41+D⋅(−2)1=4C−2D=0}⇔{C=1−DD=2C}⇔{C=1−2CD=2C}⇔{C=1/3D=2/3}\begin{array} { l } \left\{ \begin{array} { l } b _ { 0 } = C \cdot 4 ^ { 0 } + D \cdot ( - 2 ) ^ { 0 } = C + D = 1 \\b _ { 1 } = C \cdot 4 ^ { 1 } + D \cdot ( - 2 ) ^ { 1 } = 4 C - 2 D = 0\end{array} \right\} \Leftrightarrow \left\{ \begin{array} { l } C = 1 - D \\D = 2 C\end{array} \right\} \\\Leftrightarrow \left\{ \begin{array} { l } C = 1 - 2 C \\D = 2 C\end{array} \right\} \Leftrightarrow \left\{ \begin{array} { l } C = 1 / 3 \\D = 2 / 3\end{array} \right\}\end{array}
Thus bn=13⋅4n+23⋅(−2)nb _ { n } = \frac { 1 } { 3 } \cdot 4 ^ { n } + \frac { 2 } { 3 } \cdot ( - 2 ) ^ { n } for all integers n≥0n \geq 0 .

Use mathematical induction to prove that for all integers n≥1n \geq 1 , 4+8+12+⋯+4n=2n2+2n.4 + 8 + 12 + \cdots + 4 n = 2 n ^ { 2 } + 2 n .

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 Proof (by mathematical induction) ‾\underline { \text { Proof (by mathematical induction) } } : Let the property P(n)P ( n ) be the equation
4+8+12+⋯+4n=2n2+2n.←P(n)4 + 8 + 12 + \cdots + 4 n = 2 n ^ { 2 } + 2 n . \quad \leftarrow P ( n )
Note that 4+8+12+⋯+4n=∑i=1n4i4 + 8 + 12 + \cdots + 4 n = \sum _ { i = 1 } ^ { n } 4 i Show that P(1)P ( 1 ) is true: P(1)P ( 1 ) is true because the left-hand side is ∑i=114i=4⋅1=4\sum _ { i = 1 } ^ { 1 } 4 i = 4 \cdot 1 = 4 and the right-hand side is 2⋅12+2⋅1=42 \cdot 1 ^ { 2 } + 2 \cdot 1 = 4 also.

Show that for all integers k≥1k \geq 1 , if P(k)P ( k ) is true then P(k+1)P ( k + 1 ) is true: Let kk be any integer with k≥1k \geq 1 , and suppose that
4+8+12+⋯+4k=2k2+2k.←P(k) inductive hypothesis 4 + 8 + 12 + \cdots + 4 k = 2 k ^ { 2 } + 2 k . \quad \leftarrow \quad \begin{array} { l } P ( k ) \\\text { inductive hypothesis }\end{array}
We must show that
4+8+12+⋯+4(k+1)=2(k+1)2+2(k+1).←P(k+1)4 + 8 + 12 + \cdots + 4 ( k + 1 ) = 2 ( k + 1 ) ^ { 2 } + 2 ( k + 1 ) . \quad \leftarrow P ( k + 1 )
Now the left-hand side of P(k+1)P ( k + 1 ) is
4+8+12+⋯+4(k+1)=4+8+12+⋯+4k+4(k+1)= by making the next-to-last term explicit (2k2+2k)+4(k+1) by inductive hypothesis =2k2+6k+4 by multiplying out and combining like terms. \begin{aligned} 4 + 8 + 12 + \cdots + 4 ( k + 1 ) = & 4 + 8 + 12 + \cdots + 4 k + 4 ( k + 1 ) \\ = & \text { by making the next-to-last term explicit } \\ & \left( 2 k ^ { 2 } + 2 k \right) + 4 ( k + 1 ) \\ & \text { by inductive hypothesis } \\ & = \begin{array} { c } 2 k ^ { 2 } + 6 k + 4 \\ \text { by multiplying out and combining like terms. } \end{array} \end{aligned}
And the right-hand side of P(k+1)P ( k + 1 ) is
2(k+1)2+2(k+1)=2(k2+2k+1)+2k+2=2k2+4k+2+2k+2=2k2+6k+4.2 ( k + 1 ) ^ { 2 } + 2 ( k + 1 ) = 2 \left( k ^ { 2 } + 2 k + 1 \right) + 2 k + 2 = 2 k ^ { 2 } + 4 k + 2 + 2 k + 2 = 2 k ^ { 2 } + 6 k + 4 .
Thus the left-hand and right-hand sides of P(k+1)P ( k + 1 ) are equal (as was to be shown).

For each integer n≥0n \geq 0 , let P(n)P ( n ) be the equation  For each integer  n \geq 0 , let  P ( n )  be the equation   (Recall that by definition  1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { n } = \sum _ { i = 0 } ^ { n } 3 ^ { i }  ) (a) Is P(0) true? Justify your answer.  (b) In the inductive step of a proof that  P ( n )  is true for all integers  n \geq 0 , we suppose  P ( k )  is true (this is the inductive hypothesis), and then we show that  P ( k + 1 )  is true. Fill in the blanks below to write what we suppose and what we must show for this particular equation. Proof that for all integers  k \geq 0 , if  P ( k )  is true then  P ( k + 1 )  is true: Let  k  be any integer that is greater than or equal to 3 , and suppose that   We must showthat   (c) Finish the proof started in (b) above. (Recall that by definition 1+3+32+⋯+3n=∑i=0n3i1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { n } = \sum _ { i = 0 } ^ { n } 3 ^ { i } ) (a) Is P(0) true? Justify your answer. (b) In the inductive step of a proof that P(n)P ( n ) is true for all integers n≥0n \geq 0 , we suppose P(k)P ( k ) is true (this is the inductive hypothesis), and then we show that P(k+1)P ( k + 1 ) is true. Fill in the blanks below to write what we suppose and what we must show for this particular equation. Proof that for all integers k≥0k \geq 0 , if P(k)P ( k ) is true then P(k+1)P ( k + 1 ) is true: Let kk be any integer that is greater than or equal to 3 , and suppose that  For each integer  n \geq 0 , let  P ( n )  be the equation   (Recall that by definition  1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { n } = \sum _ { i = 0 } ^ { n } 3 ^ { i }  ) (a) Is P(0) true? Justify your answer.  (b) In the inductive step of a proof that  P ( n )  is true for all integers  n \geq 0 , we suppose  P ( k )  is true (this is the inductive hypothesis), and then we show that  P ( k + 1 )  is true. Fill in the blanks below to write what we suppose and what we must show for this particular equation. Proof that for all integers  k \geq 0 , if  P ( k )  is true then  P ( k + 1 )  is true: Let  k  be any integer that is greater than or equal to 3 , and suppose that   We must showthat   (c) Finish the proof started in (b) above. We must showthat For each integer  n \geq 0 , let  P ( n )  be the equation   (Recall that by definition  1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { n } = \sum _ { i = 0 } ^ { n } 3 ^ { i }  ) (a) Is P(0) true? Justify your answer.  (b) In the inductive step of a proof that  P ( n )  is true for all integers  n \geq 0 , we suppose  P ( k )  is true (this is the inductive hypothesis), and then we show that  P ( k + 1 )  is true. Fill in the blanks below to write what we suppose and what we must show for this particular equation. Proof that for all integers  k \geq 0 , if  P ( k )  is true then  P ( k + 1 )  is true: Let  k  be any integer that is greater than or equal to 3 , and suppose that   We must showthat   (c) Finish the proof started in (b) above. (c) Finish the proof started in (b) above.

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For each integer n≥0n \geq 0 , let P(n)P ( n ) be the equation
 For each integer  n \geq 0 , let  P ( n )  be the equation    (Recall that by definition  1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { n } = \sum _ { i = 0 } ^ { n } 3 ^ { i }  ) (a)  P ( 0 ) : \sum _ { i = 0 } ^ { 0 } 3 ^ { i } = \frac { 3 ^ { 0 + 1 } - 1 } { 2 }   P ( 0 )  is true because the left-hand side equals  \sum _ { i = 0 } ^ { 0 } 3 ^ { i } = 3 ^ { 0 } = 1 , and the right-hand side equals  \frac { 3 ^ { 0 + 1 } - 1 } { 2 } = \frac { 3 - 1 } { 2 } = 1  also. (b)  P ( k ) : 1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k } = \frac { 3 ^ { k + 1 } - 1 } { 2 }   P ( k + 1 ) : 1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k + 1 } = \frac { 3 ^ { ( k + 1 ) + 1 } - 1 } { 2 } ,  Or, equivalently,  P ( k + 1 )  is  1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k + 1 } = \frac { 3 ^ { k + 2 } - 1 } { 2 } . (c) Proof that for all integers  k \geq 0 , if  P ( k )  is true then  P ( k + 1 )  is true: Let  k  be any integer that is greater than or equal to 0 , and suppose that  1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k } = \frac { 3 ^ { k + 1 } - 1 } { 2 } . \quad \leftarrow \quad \begin{array} { l }  P ( k ) \\ \text { inductive hypothesis } \end{array}  We must show that  1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k + 1 } = \frac { 3 ^ { k + 2 } - 1 } { 2 } . \quad \leftarrow P ( k + 1 )  Now the left-hand side of  P ( k + 1 )  is  1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k + 1 } = 1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k } + 3 ^ { k + 1 }  by making the next-to-last term explicit  = \frac { 3 ^ { k + 1 } - 1 } { 2 } + 3 ^ { k + 1 }   = \frac { 3 ^ { k + 1 } - 1 } { 2 } + \frac { 2 \cdot 3 ^ { k + 1 } } { 2 }   = \frac { 3 \cdot 3 ^ { k + 1 } - 1 } { 2 }   = \frac { 3 ^ { k + 2 } - 1 } { 2 \mathrm { by } \text { adding the fractions and combining like terms } }  by a law of exponents,  = \frac { 3 ^ { k + 2 } - 1 } { 2 }  which equals the right-hand side of  P ( k + 1 ) . Thus the left-hand and right-hand sides of  P ( k + 1 )  are equal [as was to be shown].
(Recall that by definition 1+3+32+⋯+3n=∑i=0n3i1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { n } = \sum _ { i = 0 } ^ { n } 3 ^ { i } )
(a) P(0):∑i=003i=30+1−12P ( 0 ) : \sum _ { i = 0 } ^ { 0 } 3 ^ { i } = \frac { 3 ^ { 0 + 1 } - 1 } { 2 }
P(0)P ( 0 ) is true because the left-hand side equals ∑i=003i=30=1\sum _ { i = 0 } ^ { 0 } 3 ^ { i } = 3 ^ { 0 } = 1 , and the right-hand side equals 30+1−12=3−12=1\frac { 3 ^ { 0 + 1 } - 1 } { 2 } = \frac { 3 - 1 } { 2 } = 1 also.
(b) P(k):1+3+32+⋯+3k=3k+1−12P ( k ) : 1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k } = \frac { 3 ^ { k + 1 } - 1 } { 2 }
P(k+1):1+3+32+⋯+3k+1=3(k+1)+1−12,P ( k + 1 ) : 1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k + 1 } = \frac { 3 ^ { ( k + 1 ) + 1 } - 1 } { 2 } ,
Or, equivalently, P(k+1)P ( k + 1 ) is 1+3+32+⋯+3k+1=3k+2−121 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k + 1 } = \frac { 3 ^ { k + 2 } - 1 } { 2 } .
(c) Proof that for all integers k≥0k \geq 0 , if P(k)P ( k ) is true then P(k+1)P ( k + 1 ) is true:
Let kk be any integer that is greater than or equal to 0 , and suppose that
1+3+32+⋯+3k=3k+1−12.←P(k) inductive hypothesis 1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k } = \frac { 3 ^ { k + 1 } - 1 } { 2 } . \quad \leftarrow \quad \begin{array} { l } P ( k ) \\\text { inductive hypothesis }\end{array}
We must show that
1+3+32+⋯+3k+1=3k+2−12.←P(k+1)1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k + 1 } = \frac { 3 ^ { k + 2 } - 1 } { 2 } . \quad \leftarrow P ( k + 1 )
Now the left-hand side of P(k+1)P ( k + 1 ) is
1+3+32+⋯+3k+1=1+3+32+⋯+3k+3k+11 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k + 1 } = 1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { k } + 3 ^ { k + 1 }
by making the next-to-last term explicit
=3k+1−12+3k+1= \frac { 3 ^ { k + 1 } - 1 } { 2 } + 3 ^ { k + 1 }
=3k+1−12+2⋅3k+12= \frac { 3 ^ { k + 1 } - 1 } { 2 } + \frac { 2 \cdot 3 ^ { k + 1 } } { 2 } =3⋅3k+1−12= \frac { 3 \cdot 3 ^ { k + 1 } - 1 } { 2 } =3k+2−12by adding the fractions and combining like terms = \frac { 3 ^ { k + 2 } - 1 } { 2 \mathrm { by } \text { adding the fractions and combining like terms } } by a law of exponents, =3k+2−12= \frac { 3 ^ { k + 2 } - 1 } { 2 }
which equals the right-hand side of P(k+1)P ( k + 1 ) .
Thus the left-hand and right-hand sides of P(k+1)P ( k + 1 ) are equal [as was to be shown].

Use mathematical induction to prove that for all integers n≥0n \geq 0 , 1+3+32+⋯+3n=3n+1−12.1 + 3 + 3 ^ { 2 } + \cdots + 3 ^ { n } = \frac { 3 ^ { n + 1 } - 1 } { 2 } .

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A sequence S1,S2,S3S _ { 1 } , S _ { 2 } , S _ { 3 } ... is defined recursively as follows: =5+ for all integers k\geq3 =4 =8 Use (strong) mathematical induction to prove that sn is divisible by 4 for all integers n≥1n \geq 1

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Use mathematical induction to prove that for all integers n≥0,8n−1 is divisible by 7. n \geq 0,8 ^ { n } - 1 \text { is divisible by } 7 \text {. }

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A sequence is defined recursively as follows: sk=5sk−1+1s _ { k } = 5 s _ { k - 1 } + 1 for all integers k≥1k \geq 1 s0=1. s _ { 0 } = 1 \text {. } Use mathematical induction to verify that this sequence satisfies the explicit formula sn=5n+1−14 for all integers n≥0. s _ { n } = \frac { 5 ^ { n + 1 } - 1 } { 4 } \quad \text { for all integers } n \geq 0 \text {. }

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For each integer n≥3n \geq 3 , let P(n)P ( n ) be the equation 3+4+5+⋯+n=(n−2)(n+3)2⋅←P(n)3+4+5+\cdots+n=\frac{(n-2)(n+3)}{2} \cdot \leftarrow P(n) (Recall that by definition 3+4+5+⋯+n=∑i=3ni.3 + 4 + 5 + \cdots + n = \sum _ { i = 3 } ^ { n } i _ {. } ) (a) Is P(3)P ( 3 ) true? Justify your answer. (b) In the inductive step of a proof that P(n)P ( n ) is true for all integers n≥3n \geq 3 , we suppose P(k)P ( k ) is true (this is the inductive hypothesis), and then we show that P(k+1)P ( k + 1 ) is true. Fill in the blanks below to write what we suppose and what we must show for this particular equation. Proof that for all integers k≥3k \geq 3 , if P(k)P ( k ) is true then P(k+1)P ( k + 1 ) is true: Let kk be any integer that is greater than or equal to 3 , and suppose that___ We must show that________ (c) Finish the proof started in (b) above.

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A sequence d1,d2,d3,…d _ { 1 } , d _ { 2 } , d _ { 3 } , \ldots satisfies the recurrence relation dk=8dk−1−16dk−2d _ { k } = 8 d _ { k - 1 } - 16 d _ { k - 2 } with initial conditions d1=0d _ { 1 } = 0 and d2=1d _ { 2 } = 1 . Find an explicit formula for the sequence.

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A sequence c0,c1,c2,…c _ { 0 } , c _ { 1 } , c _ { 2 } , \ldots satisfies the recurrence relation ck=6ck−1−9ck−2c _ { k } = 6 c _ { k - 1 } - 9 c _ { k - 2 } with initial conditions c0=1c _ { 0 } = 1 and c1=6c _ { 1 } = 6 . Find an explicit formula for the sequence.

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Use mathematical induction to prove that for all integers n≥3n \geq 3 , 2⋅3+3⋅4+⋯+(n−1)⋅n=(n−2)(n2+2n+3)3.2 \cdot 3 + 3 \cdot 4 + \cdots + ( n - 1 ) \cdot n = \frac { ( n - 2 ) \left( n ^ { 2 } + 2 n + 3 \right) } { 3 } .

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A sequence a0,a1,a2,…a _ { 0 } , a _ { 1 } , a _ { 2 } , \ldots is defined recursively as follows: =2,=9 =5-6 for all integers k\geq2 Use strong mathematical induction to prove that for all integers n≥0,an=5⋅3n−3⋅2nn \geq 0 , a _ { n } = 5 \cdot 3 ^ { n } - 3 \cdot 2 ^ { n }

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Use mathematical induction to prove that for all integers n≥5,1+4n<2nn \geq 5,1 + 4 n < 2 ^ { n }

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Use mathematical induction to prove that for all integers n≥3n \geq 3 , 3+4+5+⋯+n=(n−2)(n+3)2.3 + 4 + 5 + \cdots + n = \frac { ( n - 2 ) ( n + 3 ) } { 2 } .

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Use summation notation to rewrite the following: 13−23+33−43+531 ^ { 3 } - 2 ^ { 3 } + 3 ^ { 3 } - 4 ^ { 3 } + 5 ^ { 3 }

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For each integer n≥3n \geq 3 , let P(n)P ( n ) be the equation  For each integer  n \geq 3 , let  P ( n )  be the equation   (Recall that by definition  \left. 2 \cdot 3 + 3 \cdot 4 + \cdots + ( n - 1 ) \cdot n = \sum _ { i = 3 } ^ { n } ( i - 1 ) \cdot i . \right)  (a) Is  P ( 3 )  true? Justify your answer. (b) In the inductive step of a proof that  P ( n )  is true for all integers  n \geq 3 , we suppose  P ( k )  is true (this is the inductive hypothesis), and then we show that  P ( k + 1 )  is true. Fill in the blanks below to write what we suppose and what we must show for this particular equation. Proof that for all integers  k \geq 3 , if  P ( k )  is true then  P ( k + 1 )  is true: Let  k  be any integer that is greater than or equal to 3 , and suppose that____ We must show that_____ (c) Finish the proof started in (b) above. (Recall that by definition 2⋅3+3⋅4+⋯+(n−1)⋅n=∑i=3n(i−1)⋅i.)\left. 2 \cdot 3 + 3 \cdot 4 + \cdots + ( n - 1 ) \cdot n = \sum _ { i = 3 } ^ { n } ( i - 1 ) \cdot i . \right) (a) Is P(3)P ( 3 ) true? Justify your answer. (b) In the inductive step of a proof that P(n)P ( n ) is true for all integers n≥3n \geq 3 , we suppose P(k)P ( k ) is true (this is the inductive hypothesis), and then we show that P(k+1)P ( k + 1 ) is true. Fill in the blanks below to write what we suppose and what we must show for this particular equation. Proof that for all integers k≥3k \geq 3 , if P(k)P ( k ) is true then P(k+1)P ( k + 1 ) is true: Let kk be any integer that is greater than or equal to 3 , and suppose that____ We must show that_____ (c) Finish the proof started in (b) above.

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A sequence c0,c1,c2,…c _ { 0 } , c _ { 1 } , c _ { 2 } , \ldots is defined as follows: c0=1 and ck=7ck−1+2 for each integer k≥1. c _ { 0 } = 1 \quad \text { and } \quad c _ { k } = 7 c _ { k - 1 } + 2 \text { for each integer } k \geq 1 \text {. } (a) Find c1c _ { 1 } and c2c _ { 2 } . (b) Simplify the expression 7n+2⋅7n−1+⋯+2⋅72+2⋅7+2.7 ^ { n } + 2 \cdot 7 ^ { n - 1 } + \cdots + 2 \cdot 7 ^ { 2 } + 2 \cdot 7 + 2 . using one of the following reference formulas: 1+2+3+⋯+n=n(n+1)21 + 2 + 3 + \cdots + n = \frac { n ( n + 1 ) } { 2 } for all integers n≥1n \geq 1 . 1+r+r2+⋯+rm=rm+1−1r−11 + r + r ^ { 2 } + \cdots + r ^ { m } = \frac { r ^ { m + 1 } - 1 } { r - 1 } for all integers m≥0m \geq 0 and all real numbers r≠1r \neq 1 . (c) Use iteration to guess an explicit formula for the sequence c0,c1,c2,…c _ { 0 } , c _ { 1 } , c _ { 2 } , \ldots

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Transform the following summation by making the change of variable j = k + 1 : ∑k=1nk2n\sum _ { k = 1 } ^ { n } \frac { k ^ { 2 } } { n }

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A sequence a1,a2,a3,…a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots is defined as follows: a1=3, and ak=4ak−1+2 for all integers k≥2. a _ { 1 } = 3 , \quad \text { and } \quad a _ { k } = 4 a _ { k - 1 } + 2 \quad \text { for all integers } k \geq 2 \text {. } (a) Find a1,a2a _ { 1 } , a _ { 2 } , and a3a _ { 3 } . (b) Supposing that a5=44⋅3+43⋅2+42⋅2+4⋅2+2a _ { 5 } = 4 ^ { 4 } \cdot 3 + 4 ^ { 3 } \cdot 2 + 4 ^ { 2 } \cdot 2 + 4 \cdot 2 + 2 , find a similar numerical expression for a6a _ { 6 } by substituting the right-hand side of this equation in place of a5a _ { 5 } in the equation a6=4⋅a5+2.a _ { 6 } = 4 \cdot a _ { 5 } + 2 . (c) Guess an explicit formula for ana _ { n } . Simplify your answer using one of the following reference formulas: 1+2+3+⋯+n=n(n+1)21 + 2 + 3 + \cdots + n = \frac { n ( n + 1 ) } { 2 } for all integers n≥1n \geq 1 . 1+r+r2+⋯+rm=rm+1−1r−11 + r + r ^ { 2 } + \cdots + r ^ { m } = \frac { r ^ { m + 1 } - 1 } { r - 1 } for all integers m≥0m \geq 0 and all real numbers r≠1r \neq 1 .

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Define a set SS recursively as follows: I. BASIS: ϵ∈S\epsilon \in S II. RECURSION: If ss and tt are in SS , then a. 0s\inS b. s0\inS c. 1s1t\inS d. s1t1\inS III. RESTRICTION: No strings other than those derived from I and II are in SS . Use structural induction to prove that every string in SS contains an even number of l's.

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