If the sum of the first 15 terms of the serie | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given series is ${\left( {\frac{3}{4}} \right)^3} + {\left( {\frac{6}{4}} \right)^3} + {\left( {\frac{9}{4}} \right)^3} + {\left( {\frac{12}{4

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(B) The given series is ${\left( {\frac{3}{4}} ight)^3} + {\left( {\frac{6}{4}} ight)^3} + {\left( {\frac{9}{4}} ight)^3} + {\left( {\frac{12}{4}} ight)^3} + \dots$ up to $15$ terms.This can be written as $\sum_{r=1}^{15} {\left( \frac{3r}{4} ight)^3}$.Taking the constant out,we get $\left( \frac{3}{4} ight)^3 \sum_{r=1}^{15} r^3 = \frac{27}{64} \sum_{r=1}^{15} r^3$.Using the formula for the sum of cubes of the first $n$ natural numbers,$\sum_{r=1}^{n} r^3 = \left[ \frac{n(n+1)}{2} ight]^2$.For $n=15$,$\sum_{r=1}^{15} r^3 = \left[ \frac{15 	imes 16}{2} ight]^2 = (15 	imes 8)^2 = 120^2 = 14400$.Thus,the sum $S = \frac{27}{64} 	imes 14400$.$S = 27 	imes \frac{14400}{64} = 27 	imes 225$.Given that $S = 225k$,we have $225k = 27 	imes 225$.Therefore,$k = 27$.

If the sum of the first $15$ terms of the series ${\left( {\frac{3}{4}} \right)^3} + {\left( {1\frac{1}{2}} \right)^3} + {\left( {2\frac{1}{4}} \right)^3} + {3^3} + {\left( {3\frac{3}{4}} \right)^3} + \dots$ is equal to $225k$,then $k$ is equal to

Similar Questions

If the sum of the series $2 + 5 + 8 + 11 + \dots$ is $60100$, then the number of terms is:

If the sum and product of the first three terms in an $A.P.$ are $33$ and $1155$,respectively,then a value of its $11^{th}$ term is

The sum of the first $20$ terms of the sequence $0.7, 0.77, 0.777, \dots$ is

Given $x = 1 + a + a^2 + ... \infty$ $(a < 1)$ and $y = 1 + b + b^2 + ... \infty$ $(b < 1)$,find the value of $1 + ab + a^2b^2 + ... \infty$.

Let $A_n = \left( \frac{3}{4} \right) - \left( \frac{3}{4} \right)^2 + \left( \frac{3}{4} \right)^3 - \dots + (-1)^{n-1} \left( \frac{3}{4} \right)^n$ and $B_n = 1 - A_n$. Then,the least odd natural number $p$,such that $B_n > A_n$ for all $n \geq p$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module