`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))`

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.

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

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

Let U_(n)=sum_(k=1)^(n)(n)/(n^(2)+k^(2)),S_(n)=sum_(k=0)^(n-1)(n)/(n^(2)+k^(2)) then

If .^(n)C_(3)=k.n(n-1)(n-2) then k=

Let f(n)=(sum_(r=1)^(n)((1)/(r)))/(sum_(k=1)^(n)(k)/(2n-2k+1)(2n-k+1))

Let n inN and k be an integer ge0 such that S_(k)(n)=1^(k)+2^(k)+3^(k)+ . . . +n^(k) Statement-1: S_(4)(n)=(n)/(30)(n+1)(2n+1)(3n^(2)+3n+1) Statement -2: .^(k+1)C_(1)S_(k)(n)+.^(k+1)C_(2)S_(k-1)(n)+ . . . +.^(k+1)C_(k)S_(1)(n)+.^(k+1)C_(k+1)S_(0)(n)=(n+1)^(k+1)-1

Let n inN and k be an integer ge0 such that S_(k)(n)=1^(k)+2^(k)+3^(k)+ . . . +n^(k) Statement-1: S_(4)(n)=(n)/(30)(n+1)(2n+1)(3n^(2)+3n+1) Statement -2: .^(k+1)C_(1)S_(k)(n)+.^(k+1)C_(2)S_(k-1)(n)+ . . . +.^(k+1)C_(k)S_(1)(n)+.^(k+1)C_(k+1)S_(0)(n)=(n+1)^(k+1)-1

S_(n)=sum_(k=1)^(n)(k^(2)+n^(2))/(n^(3)) and T_(n)=sum_(k=0)^(n-1)(k^(2)+n^(2))/(n^(3)),n in N then

If a_(k)=(1)/(k(k+1)) for k=1,2,3, .. , n, then (sum_(k=1)^(n) a_(k))^(2)=