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

The sum of all digits of n for which sum _(r =1) ^(n ) r 2 ^(r ) = 2+2^(n+10) is :

sum_(r=1)^n r (n-r +1) is equal to :

If sum_(r=1)^n t_r= (n(n+1)(n+2))/3 then sum _(r=1)^oo 1/t_r= (A) -1 (B) 1 (C) 0 (D) none of these

If sum _(r=1)^n (r^2+1)r! =200xx201! Then n= (A) 200 (B) 201 (C) 199 (D) none of these

The value of sum_(r=1)^(n+1)(sum_(k=1)^n "^k C_(r-1)) ( where r ,k ,n in N) is equal to a. 2^(n+1)-2 b. 2^(n+1)-1 c. 2^(n+1) d. none of these

If C_r stands for .^nC_r and sum_(r=1)^n (r.C_r)/(C_(r-1)) =210 then n= (A) 19 (B) 20 (C) 21 (D) none of these

Let f(n)=sum_(r=0)^nsum_(k=r)^n(k r)dot Find the total number of divisors of f(9)dot

If sum_(r=1)^n T_r=(3^n-1), then find the sum of sum_(r=1)^n1/(T_r) .

Find sum of sum_(r=1)^n r . C (2n,r) (a) n*2^(2n-1) (b) 2^(2n-1) (c) 2^(n-1)+1 (d) None of these

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