What is the value of the following sums?
\sum_{j=3}^{7}j^3
We have the following formula:
\sum_{j=1}^{n}j^3=\dfrac{n^2\left(n+1\right)^2}{4}
We notice that:
\sum_{j=3}^{7}j^3=\sum_{j=1}^{7}j^3-\sum_{j=1}^{2}j^3
We can apply the formula twice:
\sum_{j=3}^{7}j^3=\sum_{j=1}^{7}j^3-\sum_{j=1}^{2}j^3=\dfrac{7^2\cdot8^2}{4}-\dfrac{2^2\cdot3^2}{4}=784-9=775
\sum_{j=3}^{7}j^3=775
\sum_{j=0}^{15}j^3
We have the following formula:
\sum_{j=1}^{n}j^3=\dfrac{n^2\left(n+1\right)^2}{4}
Since 0^3 adds nothing to the sum, we notice that:
\sum_{j=0}^{15}j^3=\sum_{j=1}^{15}j^3
And:
\sum_{j=1}^{15}j^3=\dfrac{15^2\left(15+1\right)^2\right)}{4}
Therefore:
\sum_{j=0}^{15}j^3=\dfrac{15^2\left(15+1\right)^2\right)}{4}=14{,}400
\sum_{j=0}^{15}j^3=14{,}400
\sum_{j=-10}^{10}j^3
We have the following formula:
\sum_{j=1}^{n}j^3=\dfrac{n^2\left(n+1\right)^2}{4}
Since adding 0^3 has no effect on the sum, we notice that:
\sum_{j=-10}^{10}j^3=\sum_{j=-10}^{-1}j^3+ \sum_{j=1}^{10}j^3
Additionally, since \left(-j\right)^3=-j^3 we have:
\sum_{j=-10}^{-1}j^2=- \sum_{j=1}^{10}j^2
Therefore:
\sum_{j=-10}^{10}j^3=\sum_{j=1}^{10}j^3-\sum_{j=1}^{10}j^3=0
\sum_{j=-10}^{10}j^3=0
\sum_{j=10}^{25}j^2
We have the following formula:
\sum_{j=1}^{n}j^3=\dfrac{n^2\left(n+1\right)^2}{4}
We notice that:
\sum_{j=10}^{25}j^3=\sum_{j=1}^{25}j^3- \sum_{j=1}^{9}j^3
We can apply the formula twice:
\sum_{j=1}^{25}j^3=\sum_{j=1}^{25}j^3- \sum_{j=1}^{9}j^3=\dfrac{25^2\cdot26^2}{4}-\dfrac{9^2\cdot10^2}{4}
Therefore:
\sum_{j=10}^{25}j^3=105{,}625-2{,}025=103{,}600
\sum_{j=10}^{25}j^3=103{,}600
\sum_{j=5}^{k_1}j^3 with k_1 \gt 5
We have the following formula:
\sum_{j=1}^{n}j^3=\dfrac{n^2\left(n+1\right)^2}{4}
We notice that:
\sum_{j=5}^{k_1}j^3=\sum_{j=1}^{k_1}j^3- \sum_{j=1}^{4}j^3
We can apply the formula twice:
\sum_{j=5}^{k_1}j^3=\sum_{j=1}^{k_1}j^3- \sum_{j=1}^{4}j^3=\dfrac{k_1^2\cdot\left(k_1+1\right)^2}{4}-\dfrac{4^2\cdot5^2}{4}
Therefore:
\sum_{j=5}^{k_1}j^3=\dfrac{k_1^2\cdot\left(k_1+1\right)^2}{4}-\dfrac{4^2\cdot5^2}{4}
\sum_{j=5}^{k_1}j^3=\dfrac{k_1^2\cdot\left(k_1+1\right)^2-400}{4}
\sum_{j=-4}^{k_1}j^2 with k_1 \gt 0
We have the following formula:
\sum_{j=1}^{n}j^3=\dfrac{n^2\left(n+1\right)^2}{4}
Since the 0^3 term adds nothing to the sum, it can be removed. We notice that:
\sum_{j=-4}^{k_1}j^3=\sum_{j=1}^{k_1}j^3+ \sum_{j=-4}^{-1}j^3
Since -j^3=\left(-j\right)^3 we have:
\sum_{j=-4}^{-1}j^2=-\sum_{j=1}^{4}j^2
We can apply the formula twice:
\sum_{j=-4}^{k_1}j^3=\sum_{j=1}^{k_1}j^3- \sum_{j=1}^{4}j^3=\dfrac{k_1^2\cdot\left(k_1+1\right)^2}{4}-\dfrac{4^2\cdot5^2}{4}
Therefore:
\sum_{j=-4}^{k_1}j^3=\dfrac{k_1^2\cdot\left(k_1+1\right)^2-400}{4}
\sum_{j=-4}^{k_1}j^3=\dfrac{k_1^2\cdot\left(k_1+1\right)^2-400}{4}
\sum_{j=k_1}^{k_2}j^3 with 0 \lt k_1 \lt k_2
We have the following formula:
\sum_{j=1}^{n}j^3=\dfrac{n^2\left(n+1\right)^2}{4}
We notice that:
\sum_{j=k_1}^{k_2}j^3=\sum_{j=1}^{k_2}j^3- \sum_{j=1}^{k_1}j^3
We can apply the formula twice:
\sum_{j=k_1}^{k_2}j^3=\sum_{j=1}^{k_2}j^3- \sum_{j=1}^{k_1}j^3=\dfrac{k_2^2\left(k_2+1\right)^2}{4}-\dfrac{k_1^2\left(k_1+1\right)^2}{4}
\sum_{j=k_1}^{k_2}j^3=\dfrac{k_2^2\left(k_2+1\right)^2}{4}-\dfrac{k_1^2\left(k_1+1\right)^2}{4}