Let `S_n=(2+2)+(2^2+5)+(2^3+10)+(2^4+17)+.....` upto n brackets. If `S_n=2^(n+A)+Bn^3+cn^2+Dn+E` for all `n in N,` where `A,B,C,D,E` are constants then

Find S_n , if S_n = 1.3 + 2.4 + 3.5 + … upto n terms.

If aN = {an : n in N} and bn nn cN = dN where a, b, c, d in N and b, c are relatively prime then

The nth term of a series is given by t_n = (n^5 + n^5)/(n^4 + n^2 + 1) and if sum of its n terms can be expressed as s_n = a_n^2 + a + 1/(b_n^2 + b) , where a_n and b_n are the nth terms of some arithmetic progression and a, b are some constants , then prove that b_n/a_n is a constant.

If S_n=1^2-2^2+3^2-4^2+5^2-6^2+...... ,t h e n

S_(n) = (1+2+3+....+n)/( n) then S_(1)^(2) + S_(2)^(2) + S_(3)^(2) + ..... + S_(n)^(2) =

If S_(n)=(1.2)/(3!)+(2.2^(2))/(4!)+(3.2^(3))/(5!)+... upto n terms then the sum infinite terms is

Let S_n=1/1^2 + 1/2^2 + 1/3^2 +….. + 1/n^2 and T_n=2 -1/n , then :

Let S_n=1/1^2 + 1/2^2 + 1/3^2 +….. + 1/n^2 and T_n=2 -1/n , then :