sum_(k=1)^(20)(1+2+3+cdots+k)

Evaluate sum_(k=1)^(11)(2+3^k)

Find the value off sum_(k=1)^10(2+3^k)

The sum sum_(k=1)^(20) k (1)/(2^(k)) is equal to

The sum sum_(k=1)^(20) k (1)/(2^(k)) is equal to

sum_(k=1)^(5) (1^3 + 2^3 + ..... + k^3)/(1+3+5+..... + (2k-1) )=

Let X_(k) be real number such that X_(k)gtk^(4)+k^(2)+1 for 1 le k le 2018 . Denot N =sum_(k=1)^(2018)k . Consider the following inequalities: I. (sum_(k=1)^(2018)kx_(k))^(2)leN(sum_(k=1)^(2018)kx_(k)^(2)) " " II. (sum_(k=1)^(2018)kx_(k))^(2)leN(sum_(k=1)^(2018)k^(2)x_(k)^(2))

Evaluate sum_(k=1)^(11)(2+3^(k))

Evaluate sum_(k=1)^(11)(2+3^(k))

Evaluate :sum_(k=1)^(10)(3+2^(k))