Binomial Theorem 1 Question 17
18. The sum of coefficients of integral powers of $x$ in the binomial expansion $(1-2 \sqrt{x})^{50}$ is
(2015 Main)
(a) $\frac{1}{2}\left(3^{50}+1\right)$
(b) $\frac{1}{2}\left(3^{50}\right)$
(c) $\frac{1}{2}\left(3^{50}-1\right)$
(d) $\frac{1}{2}\left(2^{50}+1\right)$
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Answer:
Correct Answer: 18. (8)
Solution:
- Let $T _{t+1}$ be the general term in the expension of $(1-2 \sqrt{x})^{50}$
$\therefore \quad T _{r+1}={ }^{50} C _r(1)^{50 \rightarrow r}\left(-2 x^{1 / 2}\right)^{r}={ }^{50} C _r 2^{r} x^{r / 2}(-1)^{r}$
For the integral power of $x, r$ should be even integer.
$\therefore$ Sum of coefficients $=\sum _{r=0}^{25}{ }^{50} C _{2 r}(2)^{2 r}$
$$ =\frac{1}{2}\left[(1+2)^{50}+(1-2)^{50}\right]=\frac{1}{2}\left(3^{50}+1\right) $$
Alternate Solution
We have,
$$ \begin{aligned} & (1-2 \sqrt{x})^{50}=C _o-C _1 2 \sqrt{x}+C _2(\sqrt{2} x)^{2}+\ldots+C _{50}(2 \sqrt{x})^{50} \\ & (1+2 \sqrt{x})^{50}=C _o+C _1 2 \sqrt{x}+C _2(2 \sqrt{x})^{2}+\ldots+C _{50}(2 \sqrt{x})^{50} \end{aligned} $$
On adding Eqs. (i) and (ii), we get
$$ \begin{aligned} & (1-2 \sqrt{x})^{50}+(1+2 \sqrt{x})^{50} \\ & \quad=2\left[C _0+C _2(2 \sqrt{x})^{2}+\ldots+C _{50}(2 \sqrt{x})^{50}\right] \\ & \Rightarrow \quad \frac{(1-2 \sqrt{x})^{50}+(1+2 \sqrt{x})^{50}}{2} \\ & \quad=C _0+C _2(2 \sqrt{x})^{2}+\ldots+C _{50}(2 \sqrt{x})^{50} \end{aligned} $$
On putting $x=1$, we get
$$ \begin{aligned} & \frac{(1-2 \sqrt{1})^{50}+(1+2 \sqrt{1})^{50}}{2}=C _0+C _2+\ldots+C _{50}(2)^{50} \\ & \Rightarrow \quad \frac{(-1)^{50}+(3)^{50}}{2}=C _0+C _2(2)^{2}+\ldots+C _{50}(2)^{50} \\ & \Rightarrow \quad \frac{1+3^{50}}{2}=C _0+C _2(2)^{2}+\ldots+C _{50}(2)^{50} \end{aligned} $$
