Question

Find the sum of n terms of the series $\frac{1}{3}+\frac{3}{3.7}+\frac{5}{\mathrm{3.7.11}}+\frac{7}{\mathrm{3.7.11.15}}+....$

Answer 1

The $n^{th}$ term is $\frac{2n-1}{\mathrm{3.7.11....}(4n-5)(4n-1)}$.
Assume $\frac{2n-1}{\mathrm{3.7.....}(4n-5)(4n-1)}=\frac{A(n+1)+B}{\mathrm{3.7....4}n-1}-\frac{An+B}{\mathrm{3.7....}(4n-5)}$.
$\therefore 2n-1=An+(A+B)-(an+B)(4n-1)$.
On equating coefficients we have three equations involving the two unkowns a and B, and our assumption will be correct if values of A and B can be found to satisfy all three.
Equating coefficients of $n^2$, we obtain $A=0$.
Equating the absolute terms, $-1=2B;$ that is $B=-\frac{1}{2};$ and it will be found that these values of A and B satisfy the third equation.
$\therefore u_n=\frac{1}{2}\cdot \frac{1}{\mathrm{3.7.....}(4n-5)}-\frac{1}{2}\cdot \frac{1}{\mathrm{3.7....}(4n-5)(4n-1)};$
hence $S_n=\frac{1}{2}-\frac{1}{2}\cdot \frac{1}{\mathrm{3.7.11....}(4n-1)}$.