Binomial Theorem 1 Question 16
17.
If the number of terms in the expansion of $(1-\frac{2}{x}+{\frac{4}{x^{2}}})^{n}, x \neq 0$, is 28 , then the sum of the coefficients of all the terms in this expansion, is
(a) 64
(b) 2187
(c) 243
(d) 729
(2016 Main)
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Answer:
Correct Answer: 17. (d)
Solution:
- Clearly, number of terms in the expansion of
$(1-\frac{2}{x}+{\frac{4}{x^{2}}})^{n} \text { is } \frac{(n+2)(n+1)}{2} \text { or }{ }^{n+2} C_{2} $
$\text { [assuming } \frac{1}{x} \text { and } \frac{1}{x^{2}} \text { distinct] } $
$\therefore \frac{(n+2)(n+1)}{2}=28 $
$\Rightarrow (n+2)(n+1)=56=(6+1)(6+2) $
$\Rightarrow n=6$
Hence, sum of coefficients $=(1-2+4)^{6}=3^{6}=729$
Note As $\frac{1}{x}$ and $\frac{1}{x^{2}}$ are functions of same variables, therefore number of dissimilar terms will be $2 n+1$, i.e. odd, which is not possible.
Hence, it contains error.
