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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 72 of 106
Marks: +1, -0
52+62+725^2 + 6^2 + 7^2 + ..... + 20220^2
Solution:  
The given series can be written in the following way
52+62+725^2 + 6^2 + 7^2 + ..... + 20220^2 = (12+22+32+⋯+202)(1^2+2^2+3^2+\dots+20^2) - (12+22+32+42)(1^2+2^2+3^2+4^2)
= ∑k=120k2−∑k=14k2\sum\limits_{k=1}^{20} k^2 - \sum\limits_{k=1}^{4} k^2 = 20(20+1)(40+1)6\frac{20(20+1)(40+1)}{6} - 4(4+1)(8+1)6\frac{4(4+1)(8+1)}{6}
= 172206−1806\frac{17220}{6} - \frac{180}{6} = 170406\frac{17040}{6} = 2840
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