practice problems pending

Q1. Show that
(1 - 1/2^2)(1 - 1/3^2)....(1 - 1/(n+1)^2) = n+2/2n+2
S1.
Easy to show via induction.
Other way via telescopic cancellations:
Kth term is (k^2 - 1)/k^2 = (k-1)(k+1)/k^2 = (k-1)/k.(k+1)/k
Now separate out k-1/k and k+1/k terms and multiply:

2-1/2 * 3-1/3 * 4-1/4 .... n/n+1 = 1/n+1
2+1/2 * 3+1/3 ... n+2/n+1 = n+2/2
Hence proved.

Q2.
Show that:


S2.

Short method:
r*nCr = n*(n-1)C(r-1) and take sigma simply.


Longer method:
 2^(n-1) = (n-1)C0 + (n-1)C1... (n-1)C(n-1)

(n-1)Ck *n = nCk*(n-k)
How?
n * (n-1)!/k!(n-1-k)! = n! /k!(n-k-1)!
But (n-k-1)! = (n-k)!/(n-k)
So n * (n-1)!/k!(n-1-k)! = (n-k)n! /k!(n-k)! = nCk*(n-k)

So
n*2^(n-1)
= n *2^(n-1)
= n*[(n-1)C0 + (n-1)C1... (n-1)C(n-2)+ (n-1)C(n-1)]

= nC0 * (n-0) + nC1*(n-1) ... nC(n-2)*(2)+ nC(n-1)*(1)
It can be rewritten as:
= nCn*n + nC1 * (n-1) ... nC2 * 2 + nC1 * 1
= (r = 1 to n)Sigma(r*nCr)
= (r = 0 to n)Sigma(r*nCr)

Hence proved.

Q3.


S3.
Quite simple.
Let the given sum be S.
3S = 1.3^2 + 2.3^3... n.3^(n+1)
S-3S = 3 + 3^2 ... 3^n - n.3^(n+1)
And them simplify and take GP sum to get the answer.

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