SATHEE: Binomial Theorem 1 Question 16

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

(d) 729

(2016 Main)

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Answer:

Correct Answer: 17. (d)

Solution:

Clearly, number of terms in the expansion of

$(1 - \frac{2}{x} + \frac{4}{x^{2}})^{n} is \frac{(n + 2) (n + 1)}{2} or^{n + 2} C_{2}$

$[assuming \frac{1}{x} and \frac{1}{x^{2}} distinct]$

$∴ \frac{(n + 2) (n + 1)}{2} = 28$

$\Rightarrow (n + 2) (n + 1) = 56 = (6 + 1) (6 + 2)$

$\Rightarrow n = 6$

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.

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Binomial Theorem 1 Question 16

17.

Answer:

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