MCQ
જો ${(1 + ax)^n} = 1 + 8x + 24{x^2} + ....,$ તો $a$ અને $n$ મેળવો.
- ✓$2, 4$
- B$2, 3$
- C$3, 6$
- D$1, 2$
==> $1 + \frac{n}{1}ax + \frac{{n(n - 1)}}{{1.2}}{a^2}{x^2} + .... = 1 + 8x + 24{x^2} + ....$
$ \Rightarrow \,\,\,\,na = 8,\frac{{n(n - 1)}}{{1.2}}{a^2} = 24 \Rightarrow na(n - 1)a = 48$
==> $8(8 - a) = 48$==> $8 - a = 6\,\, \Rightarrow a = 2\,\, \Rightarrow n = 4$.
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