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 Sn=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    

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