The sum 1^3/1 + (1^3+2^3)/(1+3) + (1^3+2^3+3^3)/(1+3+5) + ... up to 8 terms, is:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q5 – JEE-MAIN 2026

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Home Maths Sequences and SeriesQuestion #392

The sum

\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + \dots

up to 8 terms, is:

Clue 1 of 3: Identify the general term

The problem involves a series where each term is a fraction. First, determine the general $n$ -th term of the series.

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● Video Walkthrough

Step 1: Determine the general term (

T_{n}

) of the series✦ Active

The numerator of the $n$ -th term is the sum of the first $n$ cubes, which is given by the formula $\sum_{k = 1}^{n} k^{3} = {(\frac{n (n + 1)}{2})}^{2}$ .

The denominator of the $n$ -th term is the sum of the first $n$ odd numbers, which is given by the formula $\sum_{k = 1}^{n} (2 k - 1) = n^{2}$ .

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

Step 2: Calculate the sum of the series up to 8 terms○ Expand

We need to find the sum $S = \sum_{n = 1}^{8} T_{n}$ . Substitute the expression for $T_{n}$ :

S = \sum_{n = 1}^{8} \frac{(n + 1)^{2}}{4}

Let $k = n + 1$ . When $n = 1$ , $k = 2$ . When $n = 8$ , $k = 9$ . The sum becomes:

S = \frac{1}{4} \sum_{k = 2}^{9} k^{2}

Step 3: Apply the sum of squares formula and find the final sum○ Expand

The sum of the first $N$ squares is given by $\sum_{k = 1}^{N} k^{2} = \frac{N (N + 1) (2 N + 1)}{6}$ . We can write $\sum_{k = 2}^{9} k^{2}$ as $(\sum_{k = 1}^{9} k^{2}) - 1^{2}$ .

\sum_{k = 2}^{9} k^{2} = \frac{9 (9 + 1) (2 \cdot 9 + 1)}{6} - 1^{2} = \frac{9 \cdot 10 \cdot 19}{6} - 1 = 3 \cdot 5 \cdot 19 - 1 = 285 - 1 = 284

Now, substitute this value back into the expression for $S$ :

S = \frac{1}{4} \cdot 284 = 71

💡 Teacher's Secret Hint

Ensure to subtract the $k = 1$ term when the summation starts from $k = 2$ .

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Maths Question 5 – JEE-MAIN 2026

Clue 1 of 3: Identify the general term

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