JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Sequences and Series

Rewrite the series by grouping and separating terms:

The series can be written as $(2^2-3^2+4^2-5^2+\dots +20^2-21^2)$ (first 20 terms alternating) plus $(2^2+4^2+\dots +20^2)$ (10 even squares).

Using the difference of squares pattern and the formula for sum of squares:

• The alternating part: $-(2+3+4+\dots +21)+1=-\frac{21\times 22}{2}+1=-231+1=-230$
• The additional even squares: $4(1^2+2^2+\dots +10^2)=4\times \frac{10\times 11\times 21}{6}=1540$

Total sum: $-230+1540=1310$