" Prove that "2^(k)([n],[0])([n],[k])-2^(k-1)([n],[1])([n-1],[k-1])+2^(k-2)([n],[2])([n-2],[k-2])[-......+(-1)^(k)([n],[k])([n-k],[0])=([n],[k])],[[1 pi-JEE-2003," Main ",(2,0),60]]

2 ^ (k) ([n0]) ([nk]) - 2 ^ (k-1) ([n1]) ([n-1k-1]) + 2 ^ (k-2) ([n2]) ([nk-2]) - ,,, + (- 1) ^ (k) ([nk]) ([nk]) ([n0]) ([n0])

prove that sum_(k=1)^(n)k2^(-k)=2[1-2^(-n)-n*2^(-(n+1)))

If n and k are positive integers, show that 2^k(nC 0)(n k)-2^(k-1)(nC1)(n-1Ck-1)+2^(k-2)(nC2)((n-2k-2))_dot-...+(-1)^k(nCk)+(n-kC0)=(nC k)w h e r e(n C k) stands for ^n C_kdot

If n and k are positive integers, show that 2^k( .^n C_0)(.^n C_k)-2^(k-1)(.^n C_1)(.^(n-1) C_k-1)+2^(k-2)(.^n C_2)((n-2k-2))_dot-...+ (-1)^k(^n C_k)+(.^(n-k) C_0)=(.^n C_k)w h e r e(.^n C_k) stands for .^n C_k.

Compute ((1+k)(1+k/2) ..... (1+k/n))/((1+n)(1+n/2) ..... (1+n/k))

Prove that sum_(k=1)^(n-1)(n-k)(cos(2k pi))/(n)=-(n)/(2) wheren >=3 is an integer

{:(" "Lt),(n rarr oo):} (2^(k)+4^(k)+6^(k)+....+(2n)^(k))/(n^(k+1))=

lim_(h rarr0)(n^(2)(tan^(-1)k)/(n)-k^(2)(tan^(-1)n)/(k))/(h)