The value of - 19r-120Cr-1- -1r-22r-1 is

∑ ^19_r=1^20C_r+1· ( 1)^r/2^2r+1 =1/2∑ ^19_r=1 ^20C_r+1·( 1/4)^r =1/2∑ ^20_r=2 ^20C_r·( 1/4)^r 1 = 2∑ ^20_r=2^20C_r·( 1/4)^r = 2 [ (1 1/4)^20 ( ^20C_0( 1/4)^0+ ...

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Byju's Answer

Standard XI

Mathematics

Sum of Coefficients of All Terms

The value of ...

Question

The value of $19 \sum r = 1 \frac{^{20} C_{r + 1} \cdot (- 1)^{r}}{2^{2 r + 1}}$ is

2 ({(\frac{3}{4})}^{20} + 4)

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- 2 ({(\frac{3}{4})}^{20} + 4)

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2 ({(\frac{3}{4})}^{20} - 4)

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- 2 ({(\frac{3}{4})}^{20} - 4)

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Solution

The correct option is B $- 2 ({(\frac{3}{4})}^{20} + 4)$
$19 \sum r = 1 \frac{^{20} C_{r + 1} \cdot (- 1)^{r}}{2^{2 r + 1}}$
$= \frac{1}{2}$ $19 \sum r = 1$ $^{20} C_{r + 1} \cdot {(- \frac{1}{4})}^{r}$
$= \frac{1}{2}$ $20 \sum r = 2$ $^{20} C_{r} \cdot {(- \frac{1}{4})}^{r - 1}$
$= - 2 20 \sum r = 2$ $^{20} C_{r} \cdot {(- \frac{1}{4})}^{r}$
$= - 2 [{(1 - \frac{1}{4})}^{20} - (^{20} C_{0} {(- \frac{1}{4})}^{0} +^{20} C_{1} {(- \frac{1}{4})}^{1})]$
$= - 2 ({(\frac{3}{4})}^{20} + 4)$

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Similar questions

n \sum r = 0 (- 1)^{r}^{n} C_{r} [\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \dots \infty]

is equal to

n \sum r = 0 (- 1)^{r}^{n} C_{r} [\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \dots \infty]

is equal to

Find the sum of the series
$\sum_{n}^{r = 0} (- 1)^{r}^{n} C_{r} [\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \frac{15^{r}}{2^{4 r}} \dots u p t o m t e r m s]$

Q. The value of

lim x \to \infty \frac{1 \cdot n \sum r = 1 r + 2 \cdot n - 1 \sum r = 1 r + 3 \cdot n - 2 \sum r = 1 r + . . . . + n \cdot 1}{n^{4}}

Q. If

C_{0}, C_{1}, C_{2}, \dots, C_{n}

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and

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The value of 19∑r=120Cr+1⋅(−1)r22r+1 is

The correct option is B −2((34)20+4)19∑r=120Cr+1⋅(−1)r22r+1 =1219∑r=1 20Cr+1⋅(−14)r =1220∑r=2 20Cr⋅(−14)r−1 =−220∑r=220Cr⋅(−14)r =−2[(1−14)20−(20C0(−14)0+20C1(−14)1)] =−2((34)20+4)

The value of $19 \sum r = 1 \frac{^{20} C_{r + 1} \cdot (- 1)^{r}}{2^{2 r + 1}}$ is