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

If a_1,a_2,a_3 are in G.P. having common ratio r such that sum_(k=1)^n a_(2k-1)= sum_(k=1)^na_(2k+2)!=0 then number of possible value of r is (A) 1 (B) 2 (C) 3 (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 :

If a_n=sum(r=0)^n 1/^nC_r, then sum _(r=0^n r/^nC_r equals (A) (n-1)a_n (B) na_n (C) 1/2na_n (D) none of these

sum_(r=1)^(n)(C(n,r))/(r+2) is equal to

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

If sum_(r=1)^n r=55 ,\ fin d\ sum_(r=1)^n r^3dot

If sum_(r=1)^(n)I(r)=2^(n)-1 then sum_(r=1)^(n)(1)/(I_(r)) is

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

Let f(n)=sum_(r=0)^(n)sum_(k=r)^(n)([kr]). Find the total number of divisors of f(9)