`sum_(k=1)^(2n+1)(-1)^(k-1)k^2=`

Let S_(n)=sum_(k=1)^(4n)(-1)^((k(k+1))/2)k^(2) . Then S_(n) can take values

sum_(k =1)^(n) k(1 + 1/n)^(k -1) =

Let S_(n)=sum_(k=1)^(4n)(-1)^((k(k+1))/(2))*k^(2) , then S_(n) can take value

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

If for n in N,sum_(k=0)^(2n)(-1)^(k)(2nC_(k))^(2)=A, then find the value of sum_(k=0)^(2n)(-1)^(k)(k-2n)(2nC_(k))^(2)

If a_(1),a_(2),a_(3)...a_(2n-1) are in H.P.then sum_(k=1)^(2n)(-1)^(k)(a_(k)+a_(k+1))/(a_(k)-a_(k+1)) is equal to

Evaluate the sum , sum_(k=1)^(n)(1)/(k(k+1)(k+2)……..(k+r)) .

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