[" 29."sum_(k=1)^(2n+1)(-1)^(k-1)*k^(2)=],[[(1)(n+1)(2n-1)],[(2)(n-1)(2n+1)],[(3)(n-1)(2n-1)]]

sum_(k=1)^(oo)k(1-(1)/(n))^(k-1)=a*n(n-1)bn(n+1)c*n^(2)d.(n+1)^(2)

sum_(k=1)^ook(1-1/n)^(k-1)= a. n(n-1) b. n(n+1) c. n^2 d. (n+1)^2

sum_(k=1)^ook(1-1/n)^(k-1)=>? a. n(n-1) b. n(n+1) c. n^2 d. (n+1)^2

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

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

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 ngt3, then ((n-1),(1))(a-1)^(2)-((n-1),(2))(a-2)^(2)+((n-1),(3))(a-3)^(2)-...+(-1)^(n-2)((n-1),(n-1))(a-n+1)^(2)=k then number of positvie divisors of k, (given 'a' is a prime number)

sum_(n=1)^oo 1/((n+1)(n+2)(n+3).........(n+k))