SATHEE: Binomial Theorem 1 Question 17

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) $\frac{1}{2} (3^{50} + 1)$

(b) $\frac{1}{2} (3^{50})$

(d) $\frac{1}{2} (2^{50} + 1)$

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Answer:

Correct Answer: 18. (a)

Solution:

Let $T_{t + 1}$ be the general term in the expension of $(1 - 2 \sqrt{x})^{50}$

$∴ T_{r + 1} =^{50} C_{r} (1)^{50 \to r} {(- 2 x^{1 / 2})}^{r} =^{50} C_{r} 2^{r} x^{r / 2} (- 1)^{r}$

For the integral power of $x, r$ should be even integer.

$∴$ Sum of coefficients $= \sum_{r = 0}^{25}^{50} C_{2 r} (2)^{2 r}$

$= \frac{1}{2} [(1 + 2)^{50} + (1 - 2)^{50}] = \frac{1}{2} (3^{50} + 1)$

Alternate Solution

We have,

$\begin{aligned} (1 - 2 \sqrt{x})^{50} = C_{o} - C_{1} 2 \sqrt{x} + C_{2} (\sqrt{2} x)^{2} + \dots + C_{50} (2 \sqrt{x})^{50} \dots \dots . (i) \\ (1 + 2 \sqrt{x})^{50} = C_{o} + C_{1} 2 \sqrt{x} + C_{2} (2 \sqrt{x})^{2} + \dots + C_{50} (2 \sqrt{x})^{50} \dots \dots . (i i) \end{aligned}$

On adding Eqs. (i) and (ii), we get

$\begin{aligned} (1 - 2 \sqrt{x})^{50} + (1 + 2 \sqrt{x})^{50} \\ = 2 [C_{0} + C_{2} (2 \sqrt{x})^{2} + \dots + C_{50} (2 \sqrt{x})^{50}] \dots \dots . (i i i) \\ \Rightarrow \frac{(1 - 2 \sqrt{x})^{50} + (1 + 2 \sqrt{x})^{50}}{2} \\ = C_{0} + C_{2} (2 \sqrt{x})^{2} + \dots + C_{50} (2 \sqrt{x})^{50} \end{aligned}$

On putting $x = 1$ , we get

$\begin{aligned} \frac{(1 - 2 \sqrt{1})^{50} + (1 + 2 \sqrt{1})^{50}}{2} = C_{0} + C_{2} + \dots + C_{50} (2)^{50} \\ \Rightarrow \frac{(- 1)^{50} + (3)^{50}}{2} = C_{0} + C_{2} (2)^{2} + \dots + C_{50} (2)^{50} \\ \Rightarrow \frac{1 + 3^{50}}{2} = C_{0} + C_{2} (2)^{2} + \dots + C_{50} (2)^{50} \end{aligned}$

Binomial Theorem 1 Question 17

18.

Answer:

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Binomial Theorem 1 Question 17

18.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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