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

If a_(1),a_(2),a_(3),… are in G.P. , where a_(i) in C (where C satands for set of complex numbers) having r as common ratio such that sum_(k=1)^(n)a_(2k-1)sum_(k=1)^(n)a_(2k+3) ne 0 , then the number of possible values of r is

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

The value of sum_(r=1)^(n+1)(sum_(k=1)^(n)C(k,r-1))=

If a_1,a_2,……….,a_(n+1) are in A.P. prove that sum_(k=0)^n ^nC_k.a_(k+1)=2^(n-1)(a_1+a_(n+1))

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,...,a_n be in AP whose common difference is d then prove that sum_(i=1)^n a_ia_(i+1)=n{a_1^2+na_1d+(n^2-1)/3 d^2} .

Let a_(n)=sum_(k=1)^(n)(1)/(k(n+1-k)), then for n>=2

The value of sum_(r=1)^(n+1)(sum_(k=1)^(k)C_(r-1))( where r,k,n in N) is equal to a.2^(n+1)-2b2^(n+1)-1c.2^(n+1)d. none of these

If D_k=|[1,n,n],[2k,n^2+n+2,n^2+n],[2k-1,n^2,n^2+n+2]| and sum_(k=1)^n D_k=48 ,then n equals (a)4 (b) 6 (c) 8 (d) none of these

sum_ (k = 1) ^ (n) (2 ^ (k) + 3 ^ (k-1))