The sum to `(n+1)` terms of the series `C_0/2-C_1/3+C_2/4-C_3/5+......=`

Sum to (n+1) terms of the series: (C_(0))/(2)-(C_(1))/(3)+(C_(2))/(4)-(C_(3))/(5)+ . . . is:

If C_r stands for nC_r, then the sum of first (n+1) terms of the series a C_0-(a+d)C_1+(a+2d)C_2-(a+3d)C_3+......, is

Find the sum of ( n + 1 ) terms of the following series 2 C_ 0 − 3 C_ 1 + 4 C_ 2 − 5 C_ 3 + ... ...

If C_(r) stands for nC_(r), then the sum of first (n+1) terms of the series aC_(0)-(a+d)C_(1)+(a+2d)C_(2)-(a+3d)C_(3)+...... is

Sum of the series 2C_0+C_1/2 2^2 +C_2/3 2^3+………+C_n/(n+1) 2^(n+1)

If n is odd, then sum of the series C_(0)^(2)-C_(1)^(2)+C_(2)^(2)-C_(2)^(3)+…..+(-1)^(n)C_(n)^(2) is

" The sum of the series "3*C_(0)+11*C_(1)+19*C_(2)+27*C_(3)+...+(8n+3)*C_(n)" is "