Let A be the sum of the first 20 terms and B | vedclass

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

Question

(A) The series is $a_n = n^2$ if $n$ is odd,and $a_n = 2n^2$ if $n$ is even.$A = \sum_{n=1}^{20} a_n = (1^2 + 3^2 + \dots + 19^2) + 2(2^2 + 4^2 + \dot

Vedclass · Accepted Answer

$248$

Vedclass · Answer

(A) The series is $a_n = n^2$ if $n$ is odd,and $a_n = 2n^2$ if $n$ is even.$A = \sum_{n=1}^{20} a_n = (1^2 + 3^2 + \dots + 19^2) + 2(2^2 + 4^2 + \dots + 20^2)$.$B = \sum_{n=1}^{40} a_n = (1^2 + 3^2 + \dots + 39^2) + 2(2^2 + 4^2 + \dots + 40^2)$.$B - 2A = \sum_{n=1}^{40} a_n - 2\sum_{n=1}^{20} a_n$.This can be written as $\sum_{n=21}^{40} a_n - \sum_{n=1}^{20} a_n$.For $n$ odd,$a_n = n^2$. For $n$ even,$a_n = 2n^2$.$B - 2A = \sum_{k=11}^{20} (2k-1)^2 + \sum_{k=11}^{20} 2(2k)^2 - \sum_{k=1}^{10} (2k-1)^2 - \sum_{k=1}^{10} 2(2k)^2$.Using the property $x^2 - y^2 = (x-y)(x+y)$,we get $B - 2A = \sum_{k=1}^{10} [(2k+19)^2 - (2k-1)^2] + \sum_{k=1}^{10} 2[(2k+20)^2 - (2k)^2]$.$= \sum_{k=1}^{10} (20)(4k+18) + \sum_{k=1}^{10} 2(20)(4k+20) = 20 \sum_{k=1}^{10} (4k+18 + 8k+40) = 20 \sum_{k=1}^{10} (12k+58)$.$= 20 [12 \cdot \frac{10 \cdot 11}{2} + 580] = 20 [660 + 580] = 20 [1240] = 24800$.Since $B - 2A = 100\lambda$,$100\lambda = 24800$,so $\lambda = 248$.

Let $A$ be the sum of the first $20$ terms and $B$ be the sum of the first $40$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots$. If $B - 2A = 100\lambda$,then $\lambda$ is equal to:

Similar Questions

If $1, \log_y x, \log_z y, -15 \log_x z$ are in $A.P.$,then

The sum of the series $\frac{2}{3} + \frac{8}{9} + \frac{26}{27} + \frac{80}{81} + \dots$ to $n$ terms is:

Let $S_n$ denote the sum of the first $n$ terms of an $A.P$. If $S_4 = 16$ and $S_6 = -48$,then $S_{10}$ is equal to

If twice the $11^{th}$ term of an $A.P.$ is equal to $7$ times of its $21^{st}$ term,then its $25^{th}$ term is equal to

The sixth term of an $A.P.$ is equal to $2$. The value of the common difference $x$ of the $A.P.$ which makes the product $a_1 a_4 a_5$ least is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module