First slide

Binomial theorem for positive integral Index

Question

The sum of the coefficients of all the integral powers of x in the expansion of $(1 + 2 \sqrt{x})^{80}$ is

Moderate

A

$\frac{1}{2} (3^{80} + 1)$
B

$\frac{1}{2} (3^{80} - 1)$
C

$(3^{80} + 1)$
D

$(3^{80} - 1)$

Solution

The coefficients of the integral powers of x are
$^{80} C_{0},^{80} C_{2} \cdot 2^{2},^{80} C_{4} \cdot 2^{4}, \dots,^{80} C_{80} \cdot 2^{80}$
Now, $(1 + 2)^{80} =^{80} C_{0} +^{80} C_{1} \cdot 2 +^{80} C_{2} \cdot 2^{2} + \dots +^{80} C_{80} \cdot 2^{80}$ ……. (1)
and $(1 - 2)^{80} =^{80} C_{0} -^{80} C_{1} 2 +^{80} C_{2} \cdot 2^{2} - \dots +^{80} C_{80} \cdot 2^{80}$ …….. (2)
Adding Eq. (1) and (2), we get
$3^{80} + 1 = 2 (^{80} C_{0} +^{80} C_{2} \cdot 2^{2} +^{80} C_{4} \cdot 2^{4} + \dots +^{80} C_{80} \cdot 2^{80})$
$∴^{80} C_{0} +^{80} C_{2} \cdot 2^{2} +^{80} C_{4} \cdot 2^{4} + \dots +^{80} C_{80} \cdot 2^{80} = \frac{1}{2} (3^{80} + 1)$

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