JEE Mains · Maths · STD 11 - 8. sequence and series
The positive integer n, for which the solutions of the equation \( x(x+2)+(x+2)(x+4)+....+(x+2n-2)(x+2n) = \frac{8n}{3} \) are two consecutive even integers, is :-
- A 3
- B 6
- C 12
- D 9
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
\( x(x+2)+(x+2)(x+4)+...+(x+2n-2)(x+2n)=\frac{8n}{3} \) \( \Rightarrow\sum_{r=1}^{n}(x+2r-2)(x+2r)=\frac{8n}{3} \) \( nx^{2}+2x\sum_{r=1}^{n}(2r-1)+4\sum_{r=1}^{n}r(r-1)=\frac{8n}{3} \) \( nx^{2}+2x.n^{2}+\frac{4n(n^{2}-1)}{3}-\frac{8n}{3}=0 \)…
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