Question

Find the value of $\frac{{1.2}^2+{2.3}^2+{3.4}^2+-------nterms}{1^2.+2^2.3+3^2.4+-------nterms}$.

• A. (3n+2) / (3n+3)
• B. (2n+5) / (5n+2)
• C. (4n+1) / (3n+2)
• D. (3n+5) / (3n+1)

Answer 1

The correct option is D

(3n+5) / (3n+1)

Let $T_n$ be the $n^{th}$ term of numerator and $T_n^{{}^{\prime }}$ be the $n^{th}$ term of denominator

$T_n$ = {$n^{th}$ term of series 1,2,3,.......... } $\times$ {$n^{th}termofseries{2}^2,3^2,4^2,..........$}

$T_n$ = $n(n+1{)}^2$

$T_n^{{}^{\prime }}$ = {$n^{th}termofseries{1}^2,2^2,3^2,4^2,..........$} $\times$ {$n^{th}$​ term of series 2,3,4.......... }

$T_n^{{}^{\prime }}$ = $n^2$ (n+1)

Sum of the given series = $\frac{\sum T_n}{\sum T_n^{{}^{\prime }}}$

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

= $\frac{[\frac{n(n+1)}{2}{]}^2+2[\frac{n(n+1)(2n+1)}{6}]+[\frac{n(n+1)}{2}]}{[\frac{n(n+1)}{2}{]}^2+[\frac{n(n+1)(2n+1)}{6}]}$

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

= $\frac{\frac{1}{6}(3n^2+3n+8n+4+6)}{\frac{1}{6}(3n^2+3n+4+2)}=\frac{3n^2+11n+10}{3n^2+7n+2}$
= $\frac{(3n+5)(n+2)}{(3n+1)(n+2)}=\frac{3n+5}{3n+1}$