sum_(r=1)^(20)r(2r+1)=

Find sum_(r=1)^20 (2r - 1)

The value of sum_(r=0)^(20)r(20-r)( ^(20)C_r)^2 is equal to ?

The number of divisiors of the number (sum_(r=10)^(20)5^(r-1)19C_(r-1))/(sum_(r=0)^(17)17C_(r)4^(17.26)r) is equal to

If sum_(r=1)^20(r^2+1)r!=k!20 then sum of all divisors of k of the from 7^n, n epsilin N is (A) 7 (B) 58 (C) 350 (D) none of these

sum_(r=1)^(20)(r^2)/((r!(20-r)!)^(2)) is equal to

Sum of the series sum_(r=1)^(n)(r^(2)+1)r!, is

Find the sum sum_(r=1)^(n)r(r+1)(r+2)(r+3)

sum_(r=1)^(n)r^(3)=f(n) then sum_(r=1)^(n)(2r-1)^(3) is equal to