^nC_(0)+4.^(n)C_(1)+4^(2^(n))C_(2)+..........4^(n^(n))C_(n)

The expression nC_(0)+4*nC_(1)+4^(2)nC_(2)+.....4^(n^(n))C_(n), equals

The sum of the following series upto'n' terms is given by S=^(n)C_(0)+2.^(n)C_(1)+4^(n)C_(2)+7.^(n)C_(3)+11.^(n)C_(4)+

Evaluate .^(n)C_(0).^(n)C_(2)+.^(n)C_(1).^(n)C_(3)+.^(n)C_(2).^(n)C_(4)+"...."+.^(n)C_(n-2).^(n)C_(n) .

If C_(r) stands for nC_(r), then the sum of the series (2((n)/(2))!((n)/(2))!)/(n!)[C_(0)^(2)-2C_(1)^(2)+3C_(2)^(2)-......+(-1)^(n)(n+1)C_(n)^(2)], where n is an even positive integer,is

prove that C_(0)C_(n)+C_(1)C_(n-1)+C_(2)C_(n-2)+.........+C_(n)C_(0)=2nC_(n)

Evaluate .^(n)C_(0).^(n)C_(2)+2.^(n)C_(1).^(n)C_(3)+3.^(n)C_(2).^(n)C_(4)+"...."+(n-1).^(n)C_(n-2).^(n)C_(n) .

Prove that ^nC_(0)^(2n)C_(n)-^(n)C_(1)^(2n-2)C_(n)+^(n)C_(2)^(2n-4)C_(n)-...=2^(n)

Prove that ^nC_(0)^(2n)C_(n)-^(n)C_(1)^(2n-1)C_(n)+^(n)C_(2)xx^(2n-2)C_(n)++(-1)^(n)sim nC_(n)^(n)C_(n)=1

Find the value of ^4nC_(0)+^(4n)C_(4)+^(4n)C_(8)+...+^(4n)C_(4n)