Question

The value of \[ \frac34+\frac{15}{16}+\frac{63}{64}+\cdots \text{ up to } n \text{ terms is} \]

• A. \(n-\dfrac{4^n}{3}-\dfrac13\)
• B. \(n+\dfrac{4^n}{3}-\dfrac13\)
• C. \(n+\dfrac{4^n}{3}-\dfrac13\)
• D. \(n-\dfrac{4^n}{3}+\dfrac13\)

Hint:

Rewrite series into simple standard sums.

Answer 1

The Correct Option is D

Step 1: Each term is \[ 1-\frac1{4^k} \]
Step 2: \[ S_n=n-\sum_{k=1}^n\frac1{4^k} \]
Step 3: \[ \sum_{k=1}^n\frac1{4^k}=\frac13\left(1-\frac1{4^n}\right) \]
Step 4: \[ S_n=n-\frac{4^n}{3}+\frac13 \]