Sum of the series `a^n+a^(n-1)b+^(n-2)b^2+………..+ab^n` can be obtained by taking outt `a^n or b^n` comon and using the forumula of sum of `(n+1)` terms of G.P. answer the following question:Um of coeficients of `x^50` `(1+x)^1000 +x(1+x)^999+....

Sum of the series a^n+a^(n-1)b+^(n-2)b^2+………..+ab^n can be obtained by taking outt a^n or b^n comon and using the forumula of sum of (n+1) terms of G.P. N the basis of above information answer the following question: Coefficient of x^50 in (1+x)^1000+x(1+x)^999+........+x^999(1+x)+x^1000 is (A) ^1000C_50 (B) ^1002C_50 (C) ^1001C_50 (D) ^1001C_49

Sum of series n+(n-1)x+(n-2)x^2+........+2x^(n-2)+x^(n-1).

Sum of n terms of the series (2n-1)+2(2n-3)+3(2n-5)+... is

The sum of n terms of the following series 1+(1+x)+(1+x+x^2)+.... will be

Find the sum of the series 1+2x+3x^2+(n-1)x^(n-2) using differentiation.

Find the sum of the series 1+2x+3x^2+(n-1)x^(n-2) using differentiation.

Show that the ratio of the sum of first n terms of a G.P. and the sum of (n+1)th term to (2n)th term is (1)/(r^(n)) .

If the sum of n terms of a G.P. is 3(3^(n+1))/(4^(2n)) , then find the common ratio.

Find the sum of the series (1+2x+3x^(2)+(n-1)x^(n-2)) using differentiation.

If the sum of n terms of a G.P. is 3-(3^(n+1))/(4^(2n)) , then find the common ratio.