Find the sum 1×2 + 2×3 + 3×4 + ... + n(n+1)

Question

Find the sum $S = 1 \times 2 + 2 \times 3 + 3 \times 4 + \cdots + n(n+1)$ .

Express the answer as a closed-form formula in terms of $n$ and verify for $n = 3$ .

Solution — Step by Step

The $k$ -th term of the series is $k(k+1)$ .

We need: $S_n = \sum_{k=1}^{n} k(k+1)$

Expand $k(k+1) = k^2 + k$ .

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

We use two well-known formulas:

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

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

Substituting:

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

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

S_n = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1) \cdot 2(n+2)}{6}

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

LHS = $1 \times 2 + 2 \times 3 + 3 \times 4 = 2 + 6 + 12 = 20$

RHS = $\frac{3 \times 4 \times 5}{3} = \frac{60}{3} = 20$ ✓

Why This Works

The trick in this problem is recognizing that $k(k+1)$ is a product of two consecutive integers, which expands to $k^2 + k$ . This splits the series into two standard summations that we already know formulas for.

The result $\frac{n(n+1)(n+2)}{3}$ is clean because it involves three consecutive integers $n, n+1, n+2$ . This is actually $\frac{1}{3}$ of the product $n(n+1)(n+2)$ , which relates to counting combinations: $\binom{n+2}{3} \times 2 = \frac{n(n+1)(n+2)}{3}$ .

Alternative Method

There’s an elegant “telescoping” approach using the fact that $k(k+1) = \frac{k(k+1)(k+2) - (k-1)k(k+1)}{3}$ .

Summing from $k = 1$ to $n$ , most terms cancel (telescope), and we directly get:

S_n = \frac{n(n+1)(n+2)}{3}

This avoids needing the $\sum k^2$ formula entirely, and is a powerful technique for products of consecutive integers.

This type of series (product of consecutive integers) appears in CBSE Class 11 sequences and series, and in JEE. The key pattern: $k(k+1) \to \frac{n(n+1)(n+2)}{3}$ , $k(k+1)(k+2) \to \frac{n(n+1)(n+2)(n+3)}{4}$ . Each time, the number of terms increases by 1 and the denominator increases by 1. Memorize this pattern — it speeds up calculation significantly.

Common Mistake

The most common error is forgetting to find a common denominator when adding the two standard sums. Students compute $\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$ and incorrectly add the numerators without converting to a common denominator. To add these fractions, multiply the second term by $\frac{3}{3}$ : $\frac{n(n+1)}{2} = \frac{3n(n+1)}{6}$ . Then add normally. Skipping this step gives a wrong final formula.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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