Question

Let $S_{n} = \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 :

• A. 199
• B. 99
• C. 200
• D. 19

Answer 1

The Correct Option is A

To solve the problem, we need to evaluate the expression for S_{n} given by:

S_{n} = \frac{1}{1^{3}} + \frac{1+2}{1^{3} + 2^{3}} + \frac{1+2+3}{1^{3} + 2^{3} + 3^{3}} + \ldots + \frac{1+2+\ldots+n}{1^{3} + 2^{3} +\ldots +n^{3}}

The problem states that:

100 \, S_{n} = n

We start by simplifying the general term of the series:

The numerator of the k-th term is the sum of the first k natural numbers, which is given by the formula: \frac{k(k+1)}{2}

The denominator is the sum of cubes of the first k natural numbers. This is known to be: \left( \frac{k(k+1)}{2} \right)^{2}

Substituting these into the expression for the k-th term, we get:

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

Thus, the expression for S_{n} simplifies to:

S_{n} = \sum_{k=1}^{n} \frac{2}{k(k+1)}

The sum \sum_{k=1}^{n} \frac{2}{k(k+1)} can be simplified using partial fraction decomposition:

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

This telescopes to:

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

Substituting this back into 100 \, S_{n} = n, we get:

100 \cdot \frac{2n}{n+1} = n

Solving this equation, we have:

• Multiplying both sides by n+1 gives:
• 200n = n(n+1)

Rearranging terms, we obtain:

n^{2} - 199n = 0

Factoring out n, we get:

n(n - 199) = 0

This yields solutions n = 0 or n = 199. Since n cannot be 0 in this context (as the sum should be non-zero), we conclude:

The correct value of n is 199.