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

a_(k)=(1)/(k(k+1)) for k=1,2,3,4,...n then (sum_(k=1)^(n)a_(k))^(2)=

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

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

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

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

If a_k=1/(k(k+1)) for k=1 ,2……..,n then prove that (sum_(k=1)^n a_k)^2 =n^2/(n+1)^2

If a_(k) is the coefficient of x^(k) in the expansion of (1+x+x^(2))^(n) for k=0,1,2,......,2n then a_(1)+2a_(2)+3a_(3)+...+2na_(2n)=

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

Let a_(k)=.^(n)C_(k) for 0<=k<=n and A_(k)=[[a_(k-1),00,a_(k)]] for 1<=k<=n and B=sum_(k=1)^(n-1)A_(k)*A_(k+1)=[[a,00,b]]

Evaluate : sum_(k=1)^n (2^k+3^(k-1))