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JEE Mains · Maths · STD 11 - 8. sequence and series

ધન પૂર્ણાંક n, જેના માટે સમીકરણ \( x(x+2)+(x+2)(x+4)+....+(x+2n-2)(x+2n) = \frac{8n}{3} \) ના ઉકેલો બે ક્રમિક યુગ્મ પૂર્ણાંકો છે, તે ........... છે.

  1. A 3
  2. B 6
  3. C 12
  4. D 9
Verified Solution

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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