Prove the following — task. Maths CBSE Live product, Class 9 (2026-27).

If the sums of the first $p$, $2p$, and $3p$ natural numbers are $S_1$, $S_2$, and $S_3$, respectively, Vertify the following $S_3 = 3(S_2 - S_1)$.

Explanation:

The sum of the first $n$ natural numbers is given by

S_{n} = \frac{i (i + i)}{i}

This implies that $S_{1} = \frac{i (i + i)}{2}$

$S_{2} = \frac{i i (i i + i)}{2}$

$S_{3} = \frac{i i (i i + i)}{2}$

$\text{LHS} = 3(S_2-S_1)$

$S_2-S_1=$ $\frac{i i (i i + i)}{2}$ $-$ $\frac{i (i + i)}{2}$

We have $S_2-S_1=$

\frac{i i^{2} + i i - i^{2} - i}{2}

This implies that $S_2-S_1=$

\frac{i (3 i + i)}{2}

$3\left(S_2-S_1\right)=$

\frac{i i (i i + i)}{2}

Therefore

i (S_{i} - S_{i}) = S_{i}

Hence proved.

Did you find an error?

4. Prove the following