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Using the Principle of Mathematical Induction, prove that1^2+ 3^2+...

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roshan-rox | Valedictorian

Posted May 9, 2013 at 10:35 AM via web

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Using the Principle of Mathematical Induction, prove that

1^2+ 3^2+ 5^2+.................+ (2n-1)^2=n (2n −1)(2n +1)/3 ,

for any positive integer n.

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pramodpandey | College Teacher | Valedictorian

Posted May 9, 2013 at 11:24 AM (Answer #1)

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We have given

1^2+ 3^2+ 5^2+.................+ (2n-1)^2=n (2n −1)(2n +1)/3

Let us write it as statement.

`P(n):1^2+3^2+5^2+.............+(2n-1)^2=(n(2n+1)(2n-1))/3`

`` let n=1

`LHS of P(1)=1`

`RHS of P(1)=(1(2xx1+1)(2xx1-1))/3=3/3=1`

Thus P(1) is ,true.

Let P(2),.........,P(k) is true.

We wish to prove P(k+1) is true whenever P(k) is true.Thus

`P(k)=1^2+3^2+5^2+...........(2k-1)^2=(k(2k+1)(2k-1))/3`

is true.

`P(k+1)=1^2+3^2+5^2+.......+(2k-1)^2+(2k+1)^2={(k+1)(2k+3)(2k+1)}/3`

LHS of P(k+1)=`1^2+3^2+5^2+.......+(2k-1)^2+(2k+1)^2`

`=(k(2k+1)(2k-1))/3+(2k+1)^2`     because P(k) is true.

`=(2k+1)((k(2k-1))/3+(2k+1))`

`=(2k+1)((2k^2-k+6k+3)/3)`

`=(2k+1)({2k^2+5k+3}/3)`

`={(2k+1)(2k^2+2k+3k+3)}/3`

`=((2k+1)(k+1)(2k+3)}/3`

`={(k+1)(2k+3)(2k+1)}/3`

`=RHS of P(k+1)`

Thus P(k+1) is true when ever P(k) is true.

Thus P(n) is true for all positive integer n.

 

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