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The sum of first 9 terms of the series

$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+...is:$

Option 1)
71

Option 2)
96

Option 3)
142

Option 4)
192

Answers (1)

As we learnt in

Summation of series of natural numbers -

$\sum_{k=1}^{n}K= \frac{1}{2}n\left ( n+1 \right )$

- wherein

Sum of first n natural numbers

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

and

Summation of series of natural numbers -

$\sum_{k=1}^{n}K^{3}= \frac{1}{4}n^{2}\left ( n+1 \right )^{2}$

- wherein

Sum of cubes of first n natural numbers

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

$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$

$T_{n}= \frac{1^{3}+2^{3}+3^{3}+----}{1+3+5+-----}$

$= \frac{\left ( \frac{n(n+1)}{2} \right )^{2}}{\frac{n}{2}\left [ 2\cdot 1+(n-1)\times 2 \right ]}$

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

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

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

$\therefore T_{n}= \frac{1}{4}(n^{2}+2n+1)$

$=\frac{n^{2}}{4}+\frac{n}{2}+\frac{1}{4}$

$\therefore \sum T_{n}= S_{n}= \sum \frac{n^{2}}{4}+\frac{1}{2}\sum n+\frac{1}{4}\sum1$

$= \frac{1}{4}\cdot \frac{n(n+1)(2n+1)}{6}+\frac{1}{2}\cdot \frac{n(n+1)}{2}+\frac{1}{4}\times n$

$= \frac{n(n+1)(2n+1)}{24}+\frac{n(n+1)}{4}+\frac{n}{4}$

Put n = 9 $=\frac{9\times 10\times 19}{24} +\frac{9\times 10}{4}+\frac{9}{4}$

$=\frac{3\times 5\times 19}{4} +\frac{9\times 10}{4}+\frac{9}{4}$

$=\frac{3\times 95+90+9}{4}$

$=\frac{285+99}{4}$

$=\frac{384}{4}$

$=96$

Option 1)

71

This option is incorrect

Option 2)

96

This option is correct

Option 3)

142

This option is incorrect

Option 4)

192

This option is incorrect

Posted by

Sabhrant Ambastha

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