Sum of 1 + 6 + 9(1²+2²+3²)/7 + 12(1²+2²+3²+4²)/9 + … Up to 15 Terms

JEE Main 2019. Previous Year Question.

Problem

The sum of the following series

$$1 + 6 + \frac{9(1^2+2^2+3^2)}{7} + \frac{12(1^2+2^2+3^2+4^2)}{9} + \frac{15(1^2+2^2+\cdots+5^2)}{11} + \cdots$$

up to 15 terms is:

Key insight. The first two terms, “1” and “6”, look like they don’t fit the fraction pattern of the later terms — but they actually do, if the general term formula is checked against $k=1$ and $k=2$ as well. Once every term is written in one unified formula, the sum reduces to standard $\sum k^2$ and $\sum k^3$ results.

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Approach

Identify the pattern in the $k$-th term: the coefficient, the number of squared terms in the numerator, and the denominator all follow simple rules in $k$. Simplify that general term algebraically — the sum of squares in the numerator cancels neatly against the denominator — leaving a simple polynomial in $k$ that can be summed directly.

Solution

Step 1 — Identify the general term

For the $k$-th term (starting from $k=3$, where the pattern is most visible): coefficient $=3k$, numerator $=1^2+2^2+\cdots+k^2 = \dfrac{k(k+1)(2k+1)}{6}$, denominator $=2k+1$. So:

$$t_k = \frac{3k \cdot \dfrac{k(k+1)(2k+1)}{6}}{2k+1} = \frac{3k \cdot k(k+1)}{6} = \frac{k^2(k+1)}{2}$$

Step 2 — Check the formula against the first two terms

For $k=1$: $t_1 = \dfrac{1^2 \times 2}{2} = 1$ — matches the given first term.

For $k=2$: $t_2 = \dfrac{2^2 \times 3}{2} = 6$ — matches the given second term.

So the entire series, from $k=1$ to $k=15$, is described by the single formula $t_k = \dfrac{k^2(k+1)}{2}$, with no special cases needed.

Step 3 — Expand and split the sum

$$\sum_{k=1}^{15} t_k = \sum_{k=1}^{15} \frac{k^3+k^2}{2} = \frac{1}{2}\left(\sum_{k=1}^{15}k^3 + \sum_{k=1}^{15}k^2\right)$$

Step 4 — Apply the standard summation formulas

$$\sum_{k=1}^{15}k^3 = \left(\frac{15 \times 16}{2}\right)^2 = 120^2 = 14400$$

$$\sum_{k=1}^{15}k^2 = \frac{15 \times 16 \times 31}{6} = 1240$$

Step 5 — Combine

$$\sum_{k=1}^{15} t_k = \frac{1}{2}(14400 + 1240) = \frac{15640}{2} = 7820$$

Answer

Common mistakes

• Treating “1” and “6” as separate, irregular terms rather than checking whether they already satisfy the general formula. Here they do, which considerably shortens the work — no need to sum the first two terms separately from the rest.
• Using the wrong number of terms in the sum of squares — the numerator of the $k$-th term always runs from $1^2$ up to $k^2$, not $(k-1)^2$ or $(k+1)^2$; double-checking against the $k=3$ term (three squared terms) confirms the indexing.

Practise next

• Find the sum of $t_k = \dfrac{k^2(k+1)}{2}$ up to 10 terms instead of 15, using the same $\sum k^3$ and $\sum k^2$ formulas.