If 1+(1-2^{2} cdot 1)+(1-4^{2} cdot 3)+(1-6^{ | vedclass

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

Question

(B) The given expression is $S = 1 + \sum_{n=1}^{10} (1 - (2n)^2(2n-1))$.This can be written as $S = 1 + \sum_{n=1}^{10} 1 - \sum_{n=1}^{10} (4n^2)(2n

Vedclass · Accepted Answer

$(11, 103)$

Vedclass · Answer

(B) The given expression is $S = 1 + \sum_{n=1}^{10} (1 - (2n)^2(2n-1))$.This can be written as $S = 1 + \sum_{n=1}^{10} 1 - \sum_{n=1}^{10} (4n^2)(2n-1)$.$S = 1 + 10 - 4 \sum_{n=1}^{10} (2n^3 - n^2)$.$S = 11 - 4 [2 \sum_{n=1}^{10} n^3 - \sum_{n=1}^{10} n^2]$.Using the formulas $\sum n^3 = [\frac{n(n+1)}{2}]^2$ and $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$:$S = 11 - 4 [2 \cdot (55)^2 - \frac{10 \cdot 11 \cdot 21}{6}]$.$S = 11 - 4 [2 \cdot 3025 - 385] = 11 - 4 [6050 - 385] = 11 - 4 [5665]$.$S = 11 - 22660 = 11 - 220(103)$.Comparing this with $\alpha - 220 \beta$,we get $\alpha = 11$ and $\beta = 103$.Thus,the ordered pair is $(11, 103)$.

If $1+(1-2^{2} \cdot 1)+(1-4^{2} \cdot 3)+(1-6^{2} \cdot 5)+\ldots+(1-20^{2} \cdot 19) = \alpha - 220 \beta$,then the ordered pair $(\alpha, \beta)$ is equal to:

Similar Questions

The value of the expression $1.(2 - \omega )(2 - {\omega ^2}) + 2.(3 - \omega )(3 - {\omega ^2}) + ....... + (n - 1).(n - \omega )(n - {\omega ^2}),$ where $\omega$ is an imaginary cube root of unity,is

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = \frac{n}{n+1}$.

$2.5 + 5.9 + 8.13 + 11.17 + \ldots$ to $10$ terms $=$

The expression for $a_n$ which satisfies $a_0=0, a_1=1$ and $a_n=a_{n-1}+a_{n-2}, \forall n \in N -\{0,1\}$ is:

The sum of all terms of the $n^{th}$ bracket of the sequence $(1), (3, 5), (7, 9, 11), \dots$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module