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Questions  >>  CBSE XI  >>  Math  >>  Sequences and Series

Show that $\large\frac{1\times2^2+2\times 3^2+........n\times (n+1)^2}{1^2\times 2+2^2\times 3+.........n^2\times (n+1)}=\frac{3n+5}{3n+1}$

1 Answer

  • Sum of $n$ terms of any series $=S_n=\sum t_n$
  • $\sum n^3=\large\frac{n^2(n+1)^2}{4}$
  • $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$
  • $\sum n=\large\frac{n(n+1)}{2}$
Given series is $\large\frac{1\times2^2+2\times 3^2+........n\times (n+1)^2}{1^2\times 2+2^2\times 3+.........n^2\times (n+1)}$
Step 1
The numerator is $1\times2^2+2\times 3^2+........n\times (n+1)^2$
$n^{th}$ term of the numerator is $t_n=n(n+1)^2=n^3+2n^2+n$
We know that sum of $n$ terms of any series $=\sum t_n$
$\therefore $ Numerator$=\sum t_n=\sum(n^3+2n^2+n)$
$\qquad \qquad\:=\sum n^3+2.\sum n^2+\sum n$
We know that $\sum n^3=\large\frac{n^2(n+1)^2}{4}$, $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$ and $\sum n=\large\frac{n(n+1)}{2}$
$\Rightarrow\: $ Numerator $=\large\frac{n^2(n+1)^2}{4}$$+2.\large\frac{n(n+1)(2n+1)}{6}$$+\large\frac{n(n+1)}{2}$
Step 2
Denominator is $1^2\times 2+2^2\times 3+.........n^2\times (n+1)$
$n^{th}$ term of this series is $t_n=n^2(n+1)=n^3+n^2$
We know that sum of $n$ terms of this series is $\sum t_n$
$\therefore$ Denominator $=\sum t_n=\sum( n^3+n^2)$
$\qquad\:=\sum n^3+\sum n^2$
$\qquad \:=\large\frac{n^2(n+1)^2}{4}+\frac{n(n+1)(2n+1)}{6}$
Step 3
$\therefore$ The given series $=\large\frac{Numerator}{Denominator}$
$\qquad\: = \large\frac{n(n+1)(n+2)(3n+5)}{n(n+1)(3n+1)(n+2)}$
Hence proved.
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