How do you find the sum of the series $k (k + 2)$ $k(k+2)$ from k=1 to 4?

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How do you find the sum of the series $k (k + 2)$ $k(k+2)$ from k=1 to 4?

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Answer 1 · 2016-07-20T22:50:45

$50$ $50$

In general: $\frac{2 n^{3} + 9 n^{2} + 7 n}{6}$ $(2n^3+9n^2+7n)/6$

Explanation:

The quickest way to find the answer to this question is to add it up manually:

$4 \sum k = 1 k (k + 2) = (1 \cdot 3) + (2 \cdot 4) + (3 \cdot 5) + (4 \cdot 6)$ $sum_(k=1)^4 k(k+2) = (1*3)+(2*4)+(3*5)+(4*6)$

$= 3 + 8 + 15 + 24 = 50$ $= 3+8+15+24 = 50$

More interestingly, what would be the sum of $n$ $n$ terms for any positive integer $n$ $n$ ?

Write out the sequence of sums for $n = 1$ $n=1$ to $4$ $4$ :

$3, 11, 26, 50$ $color(blue)(3), 11, 26, 50$

Write out the sequence of differences between successive pairs of that sequence:

$8, 15, 24$ $color(blue)(8), 15, 24$

Write out the sequence of differences of that sequence:

$7, 9$ $color(blue)(7), 9$

Write out the sequence of differences of that sequence:

$2$ $color(blue)(2)$

We can now write out a formula which matches the terms of the sequence we started with, using the first element of each of these sequences as coefficients:

$s_{n} = \frac{3}{0!} + \frac{8}{1!} (n - 1) + \frac{7}{2!} (n - 1) (n - 2) + \frac{2}{3!} (n - 1) (n - 2) (n - 3)$ $s_n = color(blue)(3)/(0!) + color(blue)(8)/(1!)(n-1) + color(blue)(7)/(2!)(n-1)(n-2) + color(blue)(2)/(3!)(n-1)(n-2)(n-3)$

$= 3 + 8 n - 8 + \frac{7}{2} n^{2} - \frac{21}{2} n + 7 + \frac{1}{3} n^{3} - 2 n^{2} + \frac{11}{3} n - 2$ $=3+8n-8+7/2n^2-21/2n+7+1/3n^3-2n^2+11/3n-2$

$= \frac{2 n^{3} + 9 n^{2} + 7 n}{6}$ $=(2n^3+9n^2+7n)/6$

Note that we know that a cubic formula is sufficient to express the sum of any number of terms since the general formula for an individual term of the sequence that we are summing is quadratic.

For example:

$s_{4} = \frac{2 \cdot 4^{3} + 9 \cdot 4^{2} + 7 \cdot 4}{6} = \frac{128 + 144 + 28}{6} = \frac{300}{6} = 50$ $s_4 = (2*4^3+9*4^2+7*4)/6 = (128+144+28)/6 = 300/6 = 50$

$s_{10} = \frac{2000 + 900 + 70}{6} = \frac{2970}{6} = 495$ $s_10 = (2000+900+70)/6 = 2970/6 = 495$

Answer 2 · 2016-07-24T04:08:37

We can also do this using a couple summation formulas:

$n \sum k = 1 c x_{k} = c n \sum k = 1 x_{k}$ $sum_(k=1)^ncx_k=csum_(k=1)^nx_k" "" "$ , where $c$ $c$ is constant $" "" "$ [1]
$n \sum k = 1 (x_{k} \pm y_{k}) = n \sum k = 1 x_{k} \pm n \sum k = 1 y_{k}$ $sum_(k=1)^n(x_k+-y_k)=sum_(k=1)^nx_k+-sum_(k=1)^ny_k" "" "$ [2]
$n \sum k = 1 k = \frac{n (n + 1)}{2}$ $sum_(k=1)^nk=(n(n+1))/2" "" "$ [3]
$n \sum k = 1 k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ $sum_(k=1)^nk^2=(n(n+1)(2n+1))/6" "" "$ [4]

So, we see that:

$4 \sum k = 1 k (k + 2) = 4 \sum k = 1 (k^{2} + 2 k)$ $sum_(k=1)^4k(k+2)=sum_(k=1)^4(k^2+2k)$

Split this using [2]:

$4 \sum k = 1 (k^{2} + 2 k) = 4 \sum k = 1 k^{2} + 4 \sum k = 1 2 k$ $sum_(k=1)^4(k^2+2k)=sum_(k=1)^4k^2+sum_(k=1)^4 2k$

Rewrite the second term using [1]:

$4 \sum k = 1 k^{2} + 4 \sum k = 1 2 k = 4 \sum k = 1 k^{2} + 2 4 \sum k = 1 k$ $sum_(k=1)^4k^2+sum_(k=1)^4 2k=sum_(k=1)^4k^2+2sum_(k=1)^4k$

Now use [3] and [4] noting that $n = 4$ $n=4$ :

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

$= \frac{4 \cdot 5 \cdot 9}{6} + \frac{2 \cdot 4 \cdot 5}{2}$ $=(4*5*9)/6+(2*4*5)/2$

$= 30 + 20$ $=30+20$

$= 50$ $=50$

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