How do you find a formula for $n \sum r = 0 r 2^{r}$ $sum_(r=0)^n r2^r$ ?

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How do you find a formula for $n \sum r = 0 r 2^{r}$ $sum_(r=0)^n r2^r$ ?

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Answer 1 · 2017-01-24T22:03:12

$n \sum r = 0 r 2^{r} = (n - 1) 2^{n + 1} + 2$ $sum_(r=0)^n r2^r = (n-1)2^(n+1)+2$

Explanation:

Note that:

$(n + 1) 2^{n + 1} + n \sum r = 0 r 2^{r} = n + 1 \sum r = 0 r 2^{r}$ $(n+1)2^(n+1)+sum_(r=0)^n r2^r = sum_(r=0)^(n+1) r2^r$

$(n + 1) 2^{n + 1} + n \sum r = 0 r 2^{r} = n + 1 \sum r = 1 r 2^{r}$ $color(white)((n+1)2^(n+1)+sum_(r=0)^n r2^r) = sum_(r=1)^(n+1) r2^r$

$(n + 1) 2^{n + 1} + n \sum r = 0 r 2^{r} = n + 1 \sum r = 1 2^{r} + n + 1 \sum r = 1 (r - 1) 2^{r}$ $color(white)((n+1)2^(n+1)+sum_(r=0)^n r2^r) = sum_(r=1)^(n+1) 2^r + sum_(r=1)^(n+1) (r-1)2^r$

$(n + 1) 2^{n + 1} + n \sum r = 0 r 2^{r} = (2^{n + 2} - 2) + 2 n + 1 \sum r = 1 (r - 1) 2^{r - 1}$ $color(white)((n+1)2^(n+1)+sum_(r=0)^n r2^r) = (2^(n+2)-2) + 2sum_(r=1)^(n+1) (r-1)2^(r-1)$

$(n + 1) 2^{n + 1} + n \sum r = 0 r 2^{r} = (2^{n + 2} - 2) + 2 n \sum r = 0 r 2^{r}$ $color(white)((n+1)2^(n+1)+sum_(r=0)^n r2^r) = (2^(n+2)-2) + 2sum_(r=0)^n r2^r$

Subtract $n \sum r = 0 r 2^{r} + (2^{n + 2} - 2)$ $sum_(r=0)^n r2^r+(2^(n+2)-2)$ from both ends to get:

$n \sum r = 0 r 2^{r} = (n + 1) 2^{n + 1} - 2^{n + 2} + 2$ $sum_(r=0)^n r2^r = (n+1)2^(n+1)-2^(n+2)+2$

$n \sum r = 0 r 2^{r} = (n - 1) 2^{n + 1} + 2$ $color(white)(sum_(r=0)^n r2^r) = (n-1)2^(n+1)+2$

Answer 2 · 2017-01-25T00:50:02

$2 (1 + 2^{n} (n - 1))$ $2(1 + 2^n (n-1))$

Explanation:

$n \sum r = 0 r x^{r} = x n \sum n = 0 r x^{r - 1} = x \frac{d}{d x} (n \sum n = 0 x^{r}) =$ $sum_(r=0)^nrx^r=xsum_(n=0)^nrx^(r-1)=x d/(dx)(sum_(n=0)^nx^r)=$
$x \frac{d}{d x} (\frac{x^{n + 1} - 1}{x - 1}) = \frac{x (1 + (n (x - 1) - 1) x^{n})}{{(x - 1)}^{2}}$ $x d/(dx)((x^(n+1)-1)/(x-1))=(x(1 + (n (x-1)-1) x^(n)))/(x-1)^2$

now making $x = 2$ $x=2$ we have

$n \sum r = 0 r 2^{r} = 2 (1 + 2^{n} (n - 1))$ $sum_(r=0)^nr2^r=2(1 + 2^n (n-1))$

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How do you find a formula for $n \sum r = 0 r 2^{r}$ $sum_(r=0)^n r2^r$ ?

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