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Let

$\small \frac{1}{1^{3}} +\frac{1+2}{1^{3}+2^{3}}+\frac{1+2+3}{1^{3}+2^{3}+3^{3}}.......+\frac{1+2........+n}{1^{3}+2^{3}.......+n^{3}} .$

If 100 S_n=n, then n is equal to :

Option 1)
199

Option 2)
99

Option 3)
200

Option 4)
19

Answers (2)

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

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

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

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

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

Its $T_n$ term is $\frac{1+2+3+--n}{1^{3}+2^{3}+3^{3}+--n^{3}}$

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

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

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

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

$=2\left [ \frac{1}{n} -\frac{1}{n+1}\right ]$

$T_{1}= 2\left [ \frac{1}{1}-\frac{1}{2} \right ]$

$T_{2}= \left [ \frac{1}{2}-\frac{1}{3} \right ]$

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

$= 2\left [ 1-\frac{1}{n+1} \right ]$

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

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

Now $100S_{n}=\frac{2n}{n+1}\times100 = n (given)$

$\therefore \:n+1= 200$

$n= 199$

Option 1)

199

Correct option

Option 2)

99

Incorrect option

Option 3)

200

Incorrect option

Option 4)

19

Incorrect option

Posted by

Vakul

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