JEE Mains · Maths · STD 11 - 8. sequence and series
The sum \(1^2-2.3^2+3.5^2-4.7^2+5.9^2-\ldots +15.29^2\) is \(.......\).
- A \(6950\)
- B \(6956\)
- C \(6953\)
- D \(6952\)
Answer & Solution
Correct Answer
(D) \(6952\)
Step-by-step Solution
Detailed explanation
Separating odd placed and even placed terms we get \(S =\left(1 \cdot 1^2+3 \cdot 5^2+\ldots .15 \cdot(29)^2\right)-\left(2 \cdot 3^2+4.7^2\right.+\ldots .+14 \cdot(27)^2\) \(S =\sum \limits_{ n =1}^8(2 n -1)(4 n -3)^2-\sum_{ n =1}^7(2 n )(4 n -1)^2\) Applying summation formula…
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