The sum 1 + 1/2(1^2 + 2^2) + 1/3(1^2 + 2^2 + 3^2) + ... upto 10 terms is equal to :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q6 – JEE-MAIN 2026

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Home Maths Sequences and SeriesQuestion #400

The sum

1 + \frac{1}{2} (1^{2} + 2^{2}) + \frac{1}{3} (1^{2} + 2^{2} + 3^{2}) + \dots

upto 10 terms is equal to :

Clue 1 of 3: Identify the general term

Determine the formula for the $n$ -th term of the series, $T_{n}$ .

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Step 1: Determine the general term (

T_{n}

)✦ Active

The $n$ -th term of the series is given by $T_{n} = \frac{1}{n} (1^{2} + 2^{2} + \dots + n^{2})$ . Using the formula for the sum of the first $n$ squares, $\sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ , we can simplify $T_{n}$ .

T_{n} = \frac{1}{n} (\frac{n (n + 1) (2 n + 1)}{6}) = \frac{(n + 1) (2 n + 1)}{6} = \frac{2 n^{2} + 3 n + 1}{6}

Step 2: Calculate the sum of the series (

S_{10}

)○ Expand

The sum up to 10 terms is $S_{10} = \sum_{n = 1}^{10} T_{n}$ . Substitute the expression for $T_{n}$ and separate the summation terms.

S_{10} = \sum_{n = 1}^{10} \frac{2 n^{2} + 3 n + 1}{6} = \frac{1}{6} (2 \sum_{n = 1}^{10} n^{2} + 3 \sum_{n = 1}^{10} n + \sum_{n = 1}^{10} 1)

Step 3: Apply summation formulas and simplify○ Expand

Use the standard summation formulas for $N = 10$ : $\sum_{n = 1}^{10} n^{2} = \frac{10 (11) (21)}{6} = 385$ , $\sum_{n = 1}^{10} n = \frac{10 (11)}{2} = 55$ , and $\sum_{n = 1}^{10} 1 = 10$ . Substitute these values into the expression for $S_{10}$ and simplify.

S_{10} = \frac{1}{6} (2 (385) + 3 (55) + 10) = \frac{1}{6} (770 + 165 + 10) = \frac{1}{6} (945) = \frac{315}{2}

💡 Teacher's Secret Hint

Ensure careful calculation to avoid arithmetic errors.

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Maths Question 6 – JEE-MAIN 2026

Clue 1 of 3: Identify the general term

🧠 Full Solution Path