Maths Q5 – JEE-MAIN 2026

Home Maths Sequences and SeriesQuestion #402

The value of

1^{3} - 2^{3} + 3^{3} - 4^{3} + . . . - 14^{3} + 15^{3}

is:

Clue 1 of 3: Series Grouping

Consider grouping consecutive terms to simplify the alternating sum.

🧠 Full Solution Path

Step 1: Group Terms and Apply Difference of Cubes✦ Active

The given series is $S = 1^{3} - 2^{3} + 3^{3} - 4^{3} + . . . - 14^{3} + 15^{3}$ . We can group the terms in pairs and add the last term separately:

S = (1^{3} - 2^{3}) + (3^{3} - 4^{3}) + . . . + (13^{3} - 14^{3}) + 15^{3}

Each pair is of the form $(2 k - 1)^{3} - (2 k)^{3}$ . Using the difference of cubes formula $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ :

(2 k - 1)^{3} - (2 k)^{3} = (2 k - 1 - 2 k) ((2 k - 1)^{2} + (2 k - 1) (2 k) + (2 k)^{2})

= - 1 (4 k^{2} - 4 k + 1 + 4 k^{2} - 2 k + 4 k^{2})

= - 1 (12 k^{2} - 6 k + 1) = - 12 k^{2} + 6 k - 1

Step 2: Sum the Paired Terms○ Expand

There are 7 such pairs (from $k = 1$ to $k = 7$ ). The sum of these pairs is:

\sum_{k = 1}^{7} (- 12 k^{2} + 6 k - 1) = - 12 \sum_{k = 1}^{7} k^{2} + 6 \sum_{k = 1}^{7} k - \sum_{k = 1}^{7} 1

Using the summation formulas $\sum_{k = 1}^{n} k = \frac{n (n + 1)}{2}$ and $\sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ for $n = 7$ :

\sum_{k = 1}^{7} k = \frac{7 (8)}{2} = 28

\sum_{k = 1}^{7} k^{2} = \frac{7 (8) (15)}{6} = 7 (4) (5) = 140

Substitute these values:

S_{p a i r s} = - 12 (140) + 6 (28) - 7

S_{p a i r s} = - 1680 + 168 - 7 = - 1519

💡 Teacher's Secret Hint

Be careful with arithmetic calculations, especially with negative numbers.

Step 3: Add the Last Term to Find the Total Sum○ Expand

The last term is $15^{3}$ . Calculate its value:

15^{3} = 225 \times 15 = 3375

Now, add this to the sum of the pairs to get the total sum $S$ :

S = S_{p a i r s} + 15^{3} = - 1519 + 3375 = 1856

💡 Teacher's Secret Hint

Alternatively, the sum can be calculated as $\sum_{n = 1}^{15} n^{3} - 2 \sum_{k = 1}^{7} (2 k)^{3} = \sum_{n = 1}^{15} n^{3} - 16 \sum_{k = 1}^{7} k^{3}$ .

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Maths Question 5 – JEE-MAIN 2026

Clue 1 of 3: Series Grouping

🧠 Full Solution Path