First slide

Series of natural numbers

Question

$The value of the sum {1.2}^{2} + {2.3}^{2} + {3.4}^{2} + \dots + 19 \cdot 20^{2} is$

Moderate

A

44100
B

44000
C

41230
D

40000

Solution

${\begin{aligned} Given sum= \sum_{i = 2}^{20} (r - 1) r^{2} = \sum_{i = 1}^{20} r^{3} - \sum_{i = 1}^{20} r^{2} = (\frac{20 (21)}{2}) \\ = 44100 - 2870 = 41230 \end{aligned}}^{2} - \frac{20 (21) (41)}{6} (\sum_{i = 1}^{n} r^{3} = {(\frac{n (n + 1)}{2})}^{2}, a n d \sum_{i = 1}^{n} r^{2} = \frac{n (n + 1) (2 n + 1)}{6})$

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