Question

Show that $\frac{{1.2}^2+1.3^2+...+n}{1^2.2+2^2.3+...+n}=\frac{3n+5}{3n+1}$

Answer 1

We have to shown that

$\frac{{1.2}^2+{1.3}^2+......+n}{1^2.2+2^2.3+.....+n}=\frac{3n+5}{3n+1}$

Now

$\frac{{1.2}^2+{1.3}^2+......+n}{1^2.2+2^2.3+.....+n}=\frac{\sum _{n=1}^{n}k{\left(k+1\right)}^2}{\sum _{n=1}^{k}k\left(k+1\right)}$

$=\frac{\sum _{n=1}^n\left(k^3+2k^2+k\right)}{\sum _{n=1}^k\left(k^3+k^2\right)}$

$=\frac{\sum _{n=1}^n{k}^3+2\sum _{n=1}^n{k}^2+\sum _{n=1}^{n}k}{\sum _{n=1}^n{k}^3+\sum _{n=1}^n{k}^2}$

$=\frac{\frac{n^2{\left(n+1\right)}^2}{4}+\frac{2n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}}{\frac{n^2{\left(n+1\right)}^2}{4}+\frac{n\left(n+1\right)\left(2n+1\right)}{6}}$

$=\frac{\frac{n\left(n+1\right)}{2}\left[\frac{n+\left(n+1\right)}{2}+\frac{2.\left(2n+1\right)}{3}+1\right]}{\frac{n\left(n+1\right)}{2}\left[\frac{n+\left(n+1\right)}{2}+\frac{n\left(n+1\right)\left(2n+1\right)}{3}\right]}$

$=\frac{\frac{3n\left(n+1\right)+4\left(2n+1\right)+6}{6}}{\frac{3n\left(n+1\right)+2\left(2n+1\right)}{6}}$

$=\frac{3n^2+3n+8n+4+6}{3n^2+3n+4n+2}$

$=\frac{3n^2+11n+10}{3n^2+7n+2}$

$=\frac{3n^2+5n+6n+10}{3n^2+3n+4n+2}$

$=\frac{n\left(3n+5\right)2\left(3n+5\right)}{3n\left(n+2\right)+1\left(n+2\right)}$

$=\frac{\left(n+2\right)\left(3n+5\right)}{\left(n+2\right)\left(3n+1\right)}$

$=\frac{3n+5}{3n+1}$

Hence, we proved the given question.