What is $1 + 4 (\frac{1}{2}) + 6 {(\frac{1}{2})}^{2} + 4 {(\frac{1}{2})}^{3} + 1 {(\frac{1}{2})}^{4}$ $1+4(1/2)+6(1/2)^2+4(1/2)^3+1(1/2)^4$ ? More specifically, what is the "quick" way to solve this?

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What is $1 + 4 (\frac{1}{2}) + 6 {(\frac{1}{2})}^{2} + 4 {(\frac{1}{2})}^{3} + 1 {(\frac{1}{2})}^{4}$ $1+4(1/2)+6(1/2)^2+4(1/2)^3+1(1/2)^4$ ? More specifically, what is the "quick" way to solve this?

I believe there is a quick way for this. I just can't remember it right now.

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Answer 1 · 2017-12-15T21:12:05

$5.0625$ $5.0625$

Explanation:

Pascal's triangle:

$1$ $1$
$11$ $1 1$
$121$ $1 2 1$
$1331$ $1 3 3 1$
$14641$ $1 4 6 4 1$

the fourth row, with $14641$ $1 4 6 4 1$ , is used for binomials to the power of $4$ $4$ .

the two terms in the binomial expression are $1$ $1$ and $\frac{1}{2}$ $1/2$ .

${(1 + \frac{1}{2})}^{4} = 1^{4} \cdot 1 + 4 \cdot 1^{3} \cdot (\frac{1}{2}) + 6 \cdot 1^{2} \cdot {(\frac{1}{2})}^{2} + 4 \cdot 1^{1} {(\frac{1}{2})}^{3} + 1 {(\frac{1}{2})}^{4}$ $(1+1/2)^4 = 1^4*1 + 4*1^3*(1/2) + 6*1^2*(1/2)^2 + 4*1^1(1/2)^3 + 1(1/2)^4$

${(1 + \frac{1}{2})}^{4} = {(1 \frac{1}{2})}^{4}$ $(1+1/2)^4 = (1 1/2) ^4$

$= {1.5}^{4}$ $= 1.5^4$

$= 5.0625$ $=5.0625$

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What is $1 + 4 (\frac{1}{2}) + 6 {(\frac{1}{2})}^{2} + 4 {(\frac{1}{2})}^{3} + 1 {(\frac{1}{2})}^{4}$ $1+4(1/2)+6(1/2)^2+4(1/2)^3+1(1/2)^4$ ? More specifically, what is the "quick" way to solve this?

I believe there is a quick way for this. I just can't remember it right now.

1 Answer

Explanation:

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I believe there is a quick way for this. I just can't remember it right now.

1 Answer

Explanation:

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Impact of this question

What is $1 + 4 (\frac{1}{2}) + 6 {(\frac{1}{2})}^{2} + 4 {(\frac{1}{2})}^{3} + 1 {(\frac{1}{2})}^{4}$ $1+4(1/2)+6(1/2)^2+4(1/2)^3+1(1/2)^4$ ? More specifically, what is the "quick" way to solve this?