Question

Prove that $\frac{1}{\mathrm{1.2.3}}+\frac{5}{\mathrm{3.4.5}}+\frac{9}{\mathrm{5.6.7}}+...=\frac{p}{s}-qlo{g}_{e}t$.Find $p+s+q-t$

Answer 1

$\frac{1}{\mathrm{1.2.3}}+\frac{5}{\mathrm{3.4.5}}+\frac{9}{\mathrm{5.6.7}}+...=\frac{p}{s}-qlo{g}_{e}t$
Here, $T_n=\frac{4n-3}{(2n-1)2n(2n+1)}$
Let's resolve $\frac{4n-3}{(2n-1)2n(2n+1)}$ into partial fractions
$\frac{4n-3}{(2n-1)2n(2n+1)}=\frac{A}{(2n-1)}+\frac{B}{2n}+\frac{C}{2n+1}$
$\Rightarrow 4n-3=A(4n^2+2n)+B(4n^2-1)+C(4n^2-2n)$
$\Rightarrow 4n-3=4(A+B+C)n^2+2(A-C)n-B$
On comparing, we get
$A+B+C=0;A-C=2;B=3$
Solving , we get
$A=-\frac{1}{2},C=-\frac{5}{2}$
So, $T_n=-\frac{1}{2(2n-1)}+\frac{3}{2n}-\frac{5}{2(2n+1)}$
$=-\frac{1}{2}[(\frac{1}{2n-1}-\frac{1}{2n})-5(\frac{1}{2n}-\frac{1}{2n+1})]$
Putting n=1,2,3,4... , and addin g , we get
$S=-\frac{1}{2}[(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....)-5(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+....)]$
$\Rightarrow S=-\frac{1}{2}[(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....)+5(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+....)]$
$S=-\frac{1}{2}[\log 2+5\log 2-5]$
$\Rightarrow S=\frac{5}{2}-3\log 2$
Comparing with given value, we get
$p=\frac{5}{2},q=3,t=2$