Prove that for every integer $ n \geq 3 $,\[\left( 1 - \frac{4}{3^2} \right) \left( 1 - \frac{4}{4^2} \right) \cdots \left( 1 - \frac{4}{n^2} \right) = \frac{(n + 1)(n + 2)}{6n(n - 1)}\]

Let the given statement is we will prove the above statement by using mathematical…

Answered: Prove that for every integer n ≥ 3, (1¹…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Prove that for every integer $ n \geq 3 $,
\[
\left( 1 - \frac{4}{3^2} \right) \left( 1 - \frac{4}{4^2} \right) \cdots \left( 1 - \frac{4}{n^2} \right) = \frac{(n + 1)(n + 2)}{6n(n - 1)}
\]

Expert Solution

Step 1: first two steps in induction law:

Let the given statement is

$P left parenthesis n right parenthesis colon space open parentheses 1 minus 4 over 3 squared close parentheses open parentheses 1 minus 4 over 4 squared close parentheses........ open parentheses 1 minus 4 over n squared close parentheses space equals space fraction numerator open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses over denominator 6 n open parentheses n minus 1 close parentheses end fraction$

we will prove the above statement by using mathematical induction law as follows:

for n = 3 , L.H.S. = $open parentheses 1 minus 4 over 3 squared close parentheses space equals space 1 minus 4 over 9 space equals space 5 over 9$

R.H.S = $fraction numerator open parentheses 3 plus 1 close parentheses open parentheses 3 plus 2 close parentheses over denominator 6 open parentheses 3 close parentheses open parentheses 3 minus 1 close parentheses end fraction space equals space fraction numerator 4 cross times 5 over denominator 6 cross times 3 cross times 2 end fraction space equals space 5 over 9$

hence for n= 3 , L.H.S. = R.H.S.

that means $P left parenthesis 3 right parenthesis$ is true.

Assume that $P left parenthesis k right parenthesis$ is true for an integer $k space less than space n$ .

$P left parenthesis k right parenthesis colon space open parentheses 1 minus 4 over 3 squared close parentheses open parentheses 1 minus 4 over 4 squared close parentheses........ open parentheses 1 minus 4 over k squared close parentheses space equals space fraction numerator open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses over denominator 6 k open parentheses k minus 1 close parentheses end fraction$ ......... ( 1 )