JEE Mathematics 2023 Question with Solution

Given:

$$\frac{1^3 + 2^3 + 3^3 + \dots + n^3}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots} = \frac{9}{5}$$

Find: $n$

Use the standard summation formulas shown in the solution.

For the numerator,

$$1^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$

For the denominator, the $r$-th term is $r(2r+1)$, so

$$\sum_{r=1}^{n} r(2r+1) = \sum_{r=1}^{n} (2r^2 + r)$$ $$= 2\sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r$$

Using

$$\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}, \qquad \sum_{r=1}^{n} r = \frac{n(n+1)}{2}$$

we get

$$\sum_{r=1}^{n} (2r^2 + r) = 2 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$$ $$= \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$$ $$= \frac{n(n+1)}{6}\left(4(2n+1)+3\right)$$ $$= \frac{n(n+1)}{6}(8n+7)$$

Now substitute into the given ratio:

$$\frac{\left(\frac{n(n+1)}{2}\right)^2}{\frac{n(n+1)}{6}(8n+7)} = \frac{9}{5}$$ $$\frac{n^2(n+1)^2}{4} \cdot \frac{6}{n(n+1)(8n+7)} = \frac{9}{5}$$ $$\frac{3n(n+1)}{2(8n+7)} = \frac{9}{5}$$

Cross-multiplying,

$$5 \cdot 3n(n+1) = 9 \cdot 2(8n+7)$$ $$15n(n+1) = 18(8n+7)$$ $$15n^2 + 15n = 144n + 126$$ $$15n^2 - 129n - 126 = 0$$ $$5n^2 - 43n - 42 = 0$$ $$(5n+6)(n-5) = 0$$

So,

$$n = -\frac{6}{5} \text{ or } 5$$

Since $n$ is the number of terms, it must be a positive integer. Therefore, $n = 5$.

The required numerical value is $5$.

Given: the ratio of the sum of cubes to the series $1\cdot3 + 2\cdot5 + 3\cdot7 + \dots$ is $\frac{9}{5}$.

Find: $n$

Write the denominator termwise as

$$r(2r+1) = 2r^2 + r$$

Then

$$\sum_{r=1}^{n} r(2r+1) = 2\sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r$$

Substitute the standard results:

$$2\sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r = 2\cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$$

Combine carefully to obtain

$$\frac{n(n+1)}{6}(8n+7)$$

Now use the sum of cubes formula for the numerator:

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

Hence,

$$\frac{\left(\frac{n(n+1)}{2}\right)^2}{\frac{n(n+1)}{6}(8n+7)} = \frac{9}{5}$$

which simplifies to

$$\frac{3n(n+1)}{2(8n+7)} = \frac{9}{5}$$

After cross-multiplication and simplification,

$$5n^2 - 43n - 42 = 0$$

Factorizing,

$$(5n+6)(n-5)=0$$

Only the positive integer value is acceptable, so the answer is $5$.