# Proof by induction help please!? ((1-(1/2^2)) * (1-(1/3^2)) * 1-(1/4^2)) ... (1-(1/n^2)) = (n+1)/2nI'm specifically having trouble after the base case. Any help would be appreciated! Thanks!

### 1 Answer | Add Yours

From the given, we notice that the smallest value n can take is 2.

So when n=2, we have `1-1/(2^2)=1-1/4=3/4` and `(2+1)/[2*2]=3/4`

Hence the statement is true for n=2. Now suppose that the statement is true for n, we need to prove it for n+1. In other words we need to prove the following equality:

`(1-1/[2^2])*(1-1/[3^2])*...*(1-1/[n^2])*(1-1/[n+1]^2)=(n+1+1)/[2(n+1)]`

Let's simplify the right hand side = `(n+2)/[2(n+1)]`

Since we are assuming that the statement is true for n, then we know that `(1-1/[2^2])*(1-1/[3^2])*...*(1-1/[n^2])=(n+1)/[2n]`

If we substitue the following in the Left Hand side of the equation we are proving we get

`([n+1]/[2n])*(1-1/(n+1)^2)=`

`([n+1]/[2n])*([(n+1)^2-1]/(n+1)^2)=`

`[(n+1)((n+1)^2-1)]/[2n(n+1)^2]=`

`[(n+1)(n^2+2n+1-1)]/[2n(n+1)^2]=`

`[(n+1)(n^2+2n)]/[2n(n+1)^2]=`

`[(n+1)*n*(n+2)]/[2n(n+1)^2]=` If we simplify, we get

`(n+2)/[2(n+1)]` Which is equal to the right hand side.

Hence the statement is proved by induction.