Question

Solve the given series:

$\frac{{1.2}^2+{2.3}^2+{3.4}^2+...n(n+1{)}^2}{1.2+2^2.3+3^2.4+...n^2(n+1)}$

• A. $(3n+1)(3n+5)$
• B. $\frac{n}{2n+1}$
• C. $\frac{3n+1}{3n+5}$
• D. $\frac{3n+5}{3n+1}$

Answer 1

The correct option is D $\frac{3n+5}{3n+1}$

$=\frac{{1.2}^2+{2.3}^2+{3.4}^2+...+n(n+1{)}^2}{1^2.2+2^2.3+3^2.4+...+n^2(n+1)}$

$=\frac{(2-1){.2}^2+(3-1){.3}^2+(4-1){.4}^2+...+(n+1-1)(n+1{)}^2}{1^2.(1+1)+2^2.(2+1)+3^2.(3+1)+...+n^2(n+1)}$

$=\frac{[1^3+2^3+3^3+4^3+...+(n+1{)}^3]-[1^2+2^2+3^2+4^2+...+(n+1{)}^2]}{[1^3+2^3+3^3+...+n^3]+[1^2+2^2+3^2+...+n^2]}$

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

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

$=\frac{(n+2)[6n^2+10n]}{n[6n^2+14n+4]}$

$=\frac{2n(n+2)[3n+5]}{2n[3n^2+7n+2]}$

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

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