Sequences And Series Ques 29
- If the sum of the first $15$ terms of the series $\left(\frac{3}{4}\right)^3+\left(1 \frac{1}{2}\right)^3+\left(2 \frac{1}{4}\right)^3+3^3+\left(3 \frac{3}{4}\right)^3+\ldots$ is equal to $225$ $ k$, then $k$ is equal to
(2019 Main, 12 Jan II)
(a) $108$
(b) $27$
(c) $54$
(d) $9$
Show Answer
Answer:
Correct Answer: 29.(b)
Solution: (b) Given series is
$\left(\frac{3}{4}\right)^3+\left(1 \frac{1}{2}\right)^3+\left(2 \frac{1}{4}\right)^3+3^3+\left(3 \frac{3}{4}\right)^3+\ldots$
Let $S=\left(\frac{3}{4}\right)^3+\left(\frac{6}{4}\right)^3+\left(\frac{9}{4}\right)^3+\left(\frac{12}{4}\right)^3$ $+\left(\frac{15}{4}\right)^3+\ldots+$ upto $15$ terms
$=\left(\frac{3}{4}\right)^3\left[1^3+2^3+3^3+4^3+5^3+\ldots+15^3\right]$
$=\left(\frac{3}{4}\right)^3\left(\frac{15 \times 16}{2}\right)^2$
$\left[\because 1^3+2^3+3^3+\ldots+n^3=\left(\frac{n(n+1)}{2}\right)^2, n \in N\right]$
$=\frac{27}{64} \times \frac{225 \times 256}{4}$ $=27 \times 225$
$\Rightarrow S=27 \times 225=225 k \quad $ [given]
$\Rightarrow k=27$
BETA
