show that: (nCo)2 - (nC1)2 + (nC2)2-...(-1)n (nCn)2
=0, n is odd
(-1)n/2 (nCn/2) , n is even
here, where evr i have multiplied 2 with nC terms..their they are its power...and 'n' is also a power of -1 in the same line.
and the thing which u have to prove is having n/2 as a power of -1...
plz salve this question
62(1+x)n
(x-1)n
I think the coefficient of xn holds the key in the product.
I hoope this is good enuf.. ow let me know :)
1plzz tell in detail coz i m not able to prove it for odd and even powers of of x
62ok.. i will try..
look at another case.. this makes the solution a bit simpler!!
(1+x)n = Σ ncrxr
(1-1/x)n = Σ ncr(-1)r(1/x)r
The given series will be the coefficient of x0 in their product (see carefully!)
On the LHS we need to find the coeff of x0 in (1-1/x)n.(1+x)n
LHS= {(1-1/x)(1+x)}n
=(x-1/x)n
Now is it simple enuf? or not yet!
11term of (x-1/x)n will be
nCkxk(-1/x)n-k
=nCkx2k-n(-1)n-k
For finding coefficient of x0
We will need 2k=n no n has to be even...
Otherwise(when n is odd) the coefficient will be 0.
If n is even, k = n/2
Coefficient will be =nCn/2(-1)n-n/2
= nCn/2(-1)n/2