Let x _{1}, x _{2}, x _{3}, ldots ., x _{20} | vedclass

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

Question

$\sum x _{0}^{1}=\frac{3\left(1-\left(\frac{1}{2}\right)\right)^{20}}{1 \frac{-1}{2}}=6\left(1-\frac{1}{2^{20}}\right)$

$=\sum_{ i =1}^{20}\left( x

Vedclass · Accepted Answer

$142$

Vedclass · Answer

$\sum x _{0}^{1}=\frac{3\left(1-\left(\frac{1}{2}ight)ight)^{20}}{1 \frac{-1}{2}}=6\left(1-\frac{1}{2^{20}}ight)$

$=\sum_{ i =1}^{20}\left( x _{ i - i }ight)^{2}$

$=\sum_{ i =1}^{20}\left( x _{ i }ight)^{2}+( i )^{2}-2 x _{ i } i$

Now $=\sum_{i=1}^{20}\left(x_{i}ight)^{2}=\frac{9\left(1-\left(\frac{1}{4}ight)ight)^{20}}{1-\frac{1}{4}}=12\left(1-\frac{1}{2^{40}}ight)$

$\sum_{i=1}^{20} i^{2}=\frac{1}{6} 	imes 20 	imes 21 	imes 41=2870$

$\sum_{ i =1}^{20} x _{ i } i = s =3+2.3 \frac{1}{2}+3.3 \frac{1}{2^{2}}+4.3 \frac{1}{2^{3}}+\ldots \ldots AGP$

$=6\left(2-\frac{22}{2^{20}}ight)$

$\overline{ x }=\frac{12-\frac{12}{2^{40}}+2870-12\left(2-\frac{22}{2^{20}}ight)}{20}$

$\overline{ x }=\frac{2858}{20}+\left(\frac{-12}{2^{40}}+\frac{22}{2^{20}}ight) 	imes \frac{1}{20}$

${[\overline{ x }]=142 }$

Similar Questions

Which term of the $GP.,$ $2,8,32, \ldots$ up to $n$ terms is $131072 ?$

Find the value of $n$ so that $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ may be the geometric mean between $a$ and $b .$

The sum of $3$ numbers in geometric progression is $38$ and their product is $1728$. The middle number is

The sum of the $3^{rd}$ and the $4^{th}$ terms of a $G.P.$ is $60$ and the product of its first three terms is $1000$. If the first term of this $G.P.$ is positive, then its $7^{th}$ term is

Let $\left\{a_k\right\}$ and $\left\{b_k\right\}, k \in N$, be two G.P.s with common ratio $r_1$ and $r_2$ respectively such that $a_1=b_1=4$ and $r_1 < r_2$. Let $c_k=a_k+k, \in N$. If $c_2=5$ and $c_3=13 / 4$ then $\sum \limits_{k=1}^{\infty} c_k - \left(12 a _6+8 b _4\right)$ is equal to