The value of the infinite product 6^ (1)/(2) × 6^ (1)/(2) × 6^ (3)/(8) × 6^ (1)/(4) × is

Correct answer: B. 36 aligned & X=6^ 1 / 2 × 6^ 1 / 2 × 6^ 3 / 8 × 6^ 4 / 16 & =6^ ((1)/(2)+(2)/(4)+(3)/(8)+ ) & Let S=(1)/(2)+(2)/(4)+(3)/(8)+(4)/(16)+ ∞ & (1)/(2) S=+(1)/(4)+(2)/(8)+(3)/(16)+ ∞ & (1-(1)/(2)) S=(1)/(2)+((2-1))/(4)+((3-2))/(8)+((4-3))/(16)+ ∞ & (S)/(2)=(1)/(2)+(1)/(4)+(1)/(8)+(1)/(16)+ ∞ & (S)/(2)=((1)/(2))/(1-(1)/(2))=1. & S=2 x=6^2=36 aligned

NDA 2015

Exam: NDA
Year: 2015
Subject: Mathematics
Chapter: Sequences and Series

NDA 2015 · Mathematics · Sequences and Series

The value of the infinite product 6^ (1)/(2) × 6^ (1)/(2) × 6^...

The value of the infinite product

6^{\frac{1}{2}} \times 6^{\frac{1}{2}} \times 6^{\frac{3}{8}} \times 6^{\frac{1}{4}} \times \ldots

Options

A6
B36
C216
D $\infty$

Correct answer

B. 36

Step-by-step solution

\begin{aligned} & X=6^{1 / 2} \times 6^{1 / 2} \times 6^{3 / 8} \times 6^{4 / 16} \\ & =6^{\left(\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\ldots \ldots \ldots \ldots \ldots\right)} \\ & \text{Let } S=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\ldots \ldots \ldots \infty \\ & \frac{1}{2} S=+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\ldots \ldots \ldots \infty \\ & \left(1-\frac{1}{2}\right) S=\frac{1}{2}+\frac{(2-1)}{4}+\frac{(3-2)}{8}+\frac{(4-3)}{16}+\ldots \ldots \ldots \infty \\ & \frac{S}{2}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \ldots \ldots \ldots \infty \\ & \frac{S}{2}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1. \\ & S=2\qquad\therefore x=6^2=36\end{aligned}

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