x_i 1 2 3 n f_i 1 2^ -1 2^ -2 2^ -(n-1) What is ∑_i^n x_i f_i equal to?

Correct answer: B. 2^ n+1 -n-2 2^ n-1 x_i f_i=1.1+(2)/(2)+(3)/(2^2)+ + n 2^ n-1 ...(i) and (1)/(2) x_i f_i=(1)/(2)+(2)/(2^2)+ + n-1 2^ n-1 +(n)/(2^n) ...(ii) (i) - (ii) gives aligned & (1-(1)/(2)) x_i f_i=[1.1+(2-1)/(2)+(3-2)/(2^2)+(4-3)/(2^3)+ + 1 2^ n-1 ]-(n)/(2^n) & (1)/(2) x_t f_i=[1+(1)/(2)+(1)/(2^3)+(1)/(2^4)+ + 1 2^ n-1 ]-(n)/(2^n) aligned aligned (1)/(2) x_i f_t & =(1(1-(1)/(2^n)))/(1-(1)/(2))-(n)/(2^n) & =(2(2^n-1))/(2^n)-(n)/(2^n) x_i f_i & = 2^ n+1 -n-2 2^ n-1 aligned

NDA 2024

Exam: NDA
Year: 2024
Subject: Mathematics
Chapter: Sequences and Series

NDA 2024 · Mathematics · Sequences and Series

x_i 1 2 3 n f_i 1 2^ -1 2^ -2 2^ -(n-1) What is ∑_i^n x_i f_i equal...

x_i	1	2	3	\ldots	n
f_i	1	2^{-1}	2^{-2}	\ldots	2^{-(n-1)}

What is \sum_i^n x_i f_i

equal to?

Options

A $\frac{2^{n+1}-n+2}{2^{n-1}}$
B $\frac{2^{n+1}-n-2}{2^{n-1}}$
C $\frac{2^{n+1}+n+2}{2^{n-1}}$
D $\frac{2^{n+1}-n-2}{2^n}$

Correct answer

B. $\frac{2^{n+1}-n-2}{2^{n-1}}$

Step-by-step solution

\because \quad \Sigma x_i f_i=1.1+\frac{2}{2}+\frac{3}{2^2}+\ldots+\frac{n}{2^{n-1}}\qquad ...(i)

and

\quad \frac{1}{2} \Sigma x_i f_i=\frac{1}{2}+\frac{2}{2^2}+\ldots+\frac{n-1}{2^{n-1}}+\frac{n}{2^n}\qquad ...(ii)

(i) - (ii) gives

\begin{aligned} & \left(1-\frac{1}{2}\right) \Sigma x_i f_i=\left[1.1+\frac{2-1}{2}+\frac{3-2}{2^2}+\frac{4-3}{2^3}+\ldots+\frac{1}{2^{n-1}}\right]-\frac{n}{2^n} \\ & \Rightarrow \frac{1}{2} \Sigma x_t f_i=\left[1+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^4}+\ldots+\frac{1}{2^{n-1}}\right]-\frac{n}{2^n} \end{aligned}

\begin{aligned} \Rightarrow \frac{1}{2} \Sigma x_i f_t & =\frac{1\left(1-\frac{1}{2^n}\right)}{1-\frac{1}{2}}-\frac{n}{2^n} \\ & =\frac{2\left(2^n-1\right)}{2^n}-\frac{n}{2^n} \\ \Rightarrow \quad \Sigma x_i f_i & =\frac{2^{n+1}-n-2}{2^{n-1}}\end{aligned}

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