First slide

Series of natural numbers

Question

$1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + \dots$ upto 22nd terms is

Moderate

A

23276
B

22736
C

22738
D

227152

Solution

$\begin{array}{l} t_{r} = 1^{2} + 2^{2} + \dots + r^{2} = \frac{1}{6} r (r + 1) (2 r + 1) \\ = \frac{1}{6} (2 r^{3} + 3 r^{2} + r) \\ ∴ \sum_{r = 1}^{n} t_{r} = \frac{1}{3} \cdot \frac{1}{4} n^{2} (n + 1)^{2} \\ + \frac{1}{2} \cdot \frac{1}{6} n (n + 1) (2 n + 1) + \frac{1}{12} n (n + 1) \\ = \frac{1}{12} n (n + 1) [n^{2} + n + 2 n + 1 + 1] \\ = \frac{1}{12} n (n + 1) (n^{2} + 3 n + 2) \\ = \frac{1}{12} n (n + 1)^{2} (n + 2) \\ ∴ \sum_{r = 1}^{22} t_{r} = \frac{1}{12} (22) (23)^{2} (24) = 23276 \end{array}$

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