First slide

Binomial theorem for positive integral Index

Question

The sum of coefficients of integral powers of r in the binomial expansion of $(1 - 2 \sqrt{x})^{50}$ is

Moderate

A

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

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

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

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

Solution

$Let T_{r + 1}$ be the general term in the expansion of $(1 - 2 \sqrt{x})^{50}$
$∴ T_{r + 1} =^{50} C_{r} (1)^{50 - r} {(- 2 x^{1 / 2})}^{r} =^{50} C_{r} 2^{r} x^{n / 2} (- 1)^{r}$
For the integral power of x, r should be even integer.
Sum of coefficients $= \sum_{r = 0}^{25}^{50} C_{2 r} (2)^{2 r} = \frac{1}{2} [(1 + 2)^{50} + (1 - 2)^{50}] = \frac{1}{2} [3^{50} + 1]$
Aliter We have,
$(1 - 2 \sqrt{x})^{50} = C_{0} - C_{1} 2 \sqrt{x} + C_{2} (2 \sqrt{x})^{2} + \dots + C_{50} (2 \sqrt{x})^{50} \dots (i) (1 + 2 \sqrt{x})^{50} = C_{0} + C_{1} 2 \sqrt{x} + C_{2} (2 \sqrt{x})^{2} + \dots + C_{50} (2 \sqrt{x})^{50} \dots (ii)$
On adding Eqs. (i) and (ii), we get
$(1 - 2 \sqrt{x})^{50} + (1 + 2 \sqrt{x})^{50} = 2 [C_{0} + C_{2} (2 \sqrt{x})^{2} + \dots + C_{50} (2 \sqrt{x})^{50}]$
$\Rightarrow \frac{(1 - 2 \sqrt{x})^{50} + (1 + 2 \sqrt{x})^{50}}{2} = C_{0} + C_{2} (2 \sqrt{x})^{2}$ $+ \dots + C_{50} (2)^{50}$
$\Rightarrow \frac{(- 1)^{50} + (3)^{50}}{2} = C_{0} + C_{2} (2)^{2} + \dots + C_{50} (2)^{50}$
$\Rightarrow \frac{1 + 3^{50}}{2} = C_{0} + C_{2} (2)^{2} + \dots + C_{50} (2)^{50}$

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