Sequences and Series: Numerical Problems Set (3)

Question

Find the sum to $n$ terms of the series $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + n(n+1)$.

Solution — Step by Step

The sum is $\dfrac{n(n+1)(n+2)}{3}$.

Why This Works

Many series with mixed terms can be split using linearity of summation. Once split, each piece is a standard sum that we know. The hard part is usually spotting the split and then simplifying the result into a clean factored form.

The compact answer $n(n+1)(n+2)/3$ has a beautiful pattern — three consecutive integers in the numerator. JEE often expects this form.

Alternative Method

Recognise that $k(k+1) = \tfrac{1}{3}[k(k+1)(k+2) - (k-1)k(k+1)]$ (telescoping). Then

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

The same answer, derived elegantly.

Common Mistake

Adding the formulas for $\sum k^2$ and $\sum k$ but botching the factoring step. Students arrive at correct unsimplified expressions but write them as the answer instead of the clean form. JEE Main marks the simplified form; CBSE accepts both.