MCQ
$1 + \frac{1}{4} + \frac{{1.3}}{{4.8}} + \frac{{1.3.5}}{{4.8.12}} + ........... = $
- ✓$\sqrt 2 $
- B$\frac{1}{{\sqrt 2 }}$
- C$\sqrt 3 $
- D$\frac{1}{{\sqrt 3 }}$
Then, $nx = \frac{1}{4}$ and $\frac{{n(n - 1)}}{2}{x^2} = \frac{1}{4}\,.\,\frac{3}{8} = \frac{3}{{32}}$
Solving these two equations for $n$ and $x$. We get $x = - \frac{1}{2}$ and $n = - \frac{1}{2}$.
$ \therefore$ Sum of the given series
= ${(1 + x)^n} = {\left( {1 - \frac{1}{2}} \right)^{ - 1/2}} = {2^{1/2}} = \sqrt {2.} $
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$y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$
અને $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ તો