Find sum to n terms of 1·2 + 2·3 + 3·4 + ... using method of differences

Question

Find the sum to $n$ terms of the series:

1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1)

(JEE Main 2021, similar pattern)

Solution — Step by Step

The $k$ th term is $T_k = k(k+1) = k^2 + k$ .

So $S_n = \sum_{k=1}^n k(k+1) = \sum_{k=1}^n (k^2 + k)$ .

S_n = \sum_{k=1}^n k^2 + \sum_{k=1}^n k

We know:

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

S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}

Factor out $\frac{n(n+1)}{6}$ :

S_n = \frac{n(n+1)}{6}\left[(2n+1) + 3\right] = \frac{n(n+1)(2n+4)}{6}

= \frac{n(n+1) \cdot 2(n+2)}{6} = \boxed{\frac{n(n+1)(n+2)}{3}}

Why This Works

The key insight is that products like $k(k+1)$ can be expanded into powers of $k$ , and we have standard formulas for $\sum k$ and $\sum k^2$ . This “expand and use standard sums” approach works for any series whose general term is a polynomial in $k$ .

The final answer $\frac{n(n+1)(n+2)}{3}$ has a beautiful combinatorial interpretation: it equals $2\binom{n+2}{3}$ , which counts something in combinatorics. This pattern — products of consecutive integers summing to binomial coefficients — appears frequently in JEE.

Alternative Method — Telescoping via Method of Differences

Write $k(k+1) = \frac{1}{3}[k(k+1)(k+2) - (k-1)k(k+1)]$ .

This is a telescoping form. Summing from $k = 1$ to $n$ :

S_n = \frac{1}{3}[n(n+1)(n+2) - 0] = \frac{n(n+1)(n+2)}{3}

Most intermediate terms cancel, leaving only the last term. This is the true “method of differences” approach.

The telescoping trick works because $k(k+1)$ is a product of 2 consecutive integers. The general rule: a product of $r$ consecutive integers can be telescoped using products of $r+1$ consecutive integers. So for $k(k+1)(k+2)$ , you’d use products of 4 consecutive integers.

Common Mistake

Students often mess up the factoring in the last step. After getting $\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$ , they try to add the fractions without taking LCM properly. The cleanest approach: factor out $\frac{n(n+1)}{6}$ first, then simplify what’s inside the brackets. Rushing the algebra here leads to answers like $\frac{n(n+1)(2n+3)}{6}$ — which is wrong but looks plausible enough to select in an MCQ.

Question

Solution — Step by Step

Why This Works

Alternative Method — Telescoping via Method of Differences

Common Mistake

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