ત્રિકોણમિતીય પ્રતિવિધેયો — ગણિત ધોરણ 12 સાયન્સ — Question

‎ $\sum n^2 = \lambda \sum n$

$\therefore \ \lambda = \frac{\sum n^2}{\sum n}$

$\therefore \ \lambda = \frac{\frac{n}{6}(n + 1) (2n + 1)}{\frac{n}{2} (n + 1)}$

$\therefore \ \lambda = \frac{2n + 1}{3}$

અહી, $sin^{-1} \left(\frac{9 \lambda^2 - 4n^2}{6 \lambda + 4n}\right)$

$= sin^{-1} \left(\frac{\frac{9 (2n + 1)^2}{9} - 4n^2}{\frac{6 (2n + 1)}{3} + 4n}\right)$

$= sin^{-1} \left(\frac{(2n + 1)^2 - 4n^2}{4n + 2 + 4n}\right)$

$= sin^{-1} \left(\frac{4n^2 + 4n + 1 - 4n^2}{8n + 2}\right)$

$= sin^{-1} \left(\frac{1}{2}\right)$

$= \frac{\pi}{6}$