Question

The sum of the series $1+\frac{9}{4}+\frac{36}{9}+\frac{100}{16}+....$ up to n terms if n=16 is

• A. 446
• B. 746
• C. 546
• D. 846

Answer 1

The correct option is A 446

The given series can be written as $1^3+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+...$
$t_n=\frac{1^3+2^3+3^3+....n^3}{1+3+5+....+(2n-1)}$
$t_n=\frac{n^2(n+1{)}^2}{4n^2}=\frac{(n+1{)}^2}{4}$
$t_n=\frac{1}{4}(n+1)(n+1)$
$=\frac{1}{4}(n^2+2n+1)=\frac{1}{4}[{\sum }_{k-1}^n{k}^2+2{\sum }_{k-1}^{n}k+n]$
$\therefore S_n=\frac{1}{4}[\frac{n(n+1)(2n+2)}{6}+n(n+1)+n]$
$\therefore S_{16}=\frac{1}{4}[\frac{\mathrm{16.17.33}}{6}+16.17+16]=\frac{1}{4}[88\times 17+16+8+16]=446$