JEE Mains · Maths · STD 11 - 8. sequence and series
The sum to \(20\) terms of the series \(2.2^2-3^2+2.4^2-5^2+2.6^2-\ldots \ldots\) is equal to \(........\).
- A \(1311\)
- B \(1312\)
- C \(1310\)
- D \(1313\)
Answer & Solution
Correct Answer
(C) \(1310\)
Step-by-step Solution
Detailed explanation
\(\left(2^2-3^2+4^2-5^2+20 \text { terms }\right)+ \left(2^2+4^2+\ldots .+10 \text { terms }\right)\) \(-(2+3+4+5+\ldots . .+11)+4\left[1+2^2+\ldots \ldots .10^2\right]\) \(-\left[\frac{21 \times 22}{2}-1\right]+4 \times \frac{10 \times 11 \times 21}{6}\)…
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