What is the value of the following sums?
\sum_{j=2}^{9}j^2
We know that:
\sum_{j=1}^{n}j^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}
Observe that:
\sum_{j=2}^{9}j^2=\sum_{j=1}^{9}j^2- \sum_{j=1}^{1}j^2
Therefore:
\sum_{j=2}^{9}j^2=\sum_{j=1}^{9}j^2- \sum_{j=1}^{1}j^2=\dfrac{9\cdot10\cdot19}{6}-\dfrac{1\cdot2\cdot3}{6}
\sum_{j=2}^{9}j^2=285-1=284
\sum_{j=2}^{9}j^2=284
\sum_{j=0}^{15}j^2
We know that:
\sum_{j=1}^{n}j^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}
Observe that:
\sum_{j=0}^{15}j^2=\sum_{j=1}^{15}j^2 since 0^2 adds nothing to the sum.
Therefore:
\sum_{j=0}^{15}j^2=\sum_{j=1}^{15}j^2=\dfrac{15\left(15+1\right)\left(2\cdot 15+1\right)}{6}
\sum_{j=0}^{15}j^2=\dfrac{\left(15\cdot 16\right) \cdot31}{6}
\sum_{j=0}^{15}j^2=1\ 240
\sum_{j=-10}^{10}j^2
We know that:
\sum_{j=1}^{n}j^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}
Observe that:
\sum_{j=-10}^{10}j^2=\sum_{j=-10}^{-1}j^2+ \sum_{j=1}^{10}j^2 since 0^2 has no effect on the sum.
Additionally,
\sum_{j=-10}^{-1}j^2= \sum_{j=1}^{10}j^2 since the square of a number is the same as the square of is opposite.
Therefore:
\sum_{j=-10}^{10}j^2=2\sum_{j=1}^{10}j^2=2\cdot\dfrac{10\cdot11\cdot21}{6}
\sum_{j=-10}^{10}j^2=770
\sum_{j=10}^{25}j^2
We know that:
\sum_{j=1}^{n}j^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}
Observe that:
\sum_{j=10}^{25}j^2=\sum_{j=1}^{25}j^2- \sum_{j=1}^{9}j^2
Therefore:
\sum_{j=1}^{25}j^2=\sum_{j=1}^{25}j^2- \sum_{j=1}^{9}j^2=\dfrac{25\cdot26\cdot51}{6}-\dfrac{9\cdot10\cdot19}{6}
\sum_{j=10}^{25}j^2=5\ 525-285
\sum_{j=10}^{25}j^2=4\ 943
\sum_{j=5}^{k_1}j^2 with k_1 \gt 0
We know that:
\sum_{j=1}^{n}j^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}
Observe that:
\sum_{j=5}^{k_1}j^2=\sum_{j=1}^{k_1}j^2- \sum_{j=1}^{4}j^2
Therefore:
\sum_{j=5}^{k_1}j^2=\sum_{j=1}^{k_1}j^2- \sum_{j=1}^{4}j^2=\dfrac{k_1\cdot\left(k_1+1\right)\cdot\left(2k_1+1\right)}{6}-\dfrac{4\cdot5\cdot9}{6}
\sum_{j=5}^{k_1}j^2=\dfrac{k_1\cdot\left(k_1+1\right)\cdot\left(2k_1+1\right)}{6}-30
\sum_{j=5}^{k_1}j^2=\dfrac{2k_1^3+3k_1^2+k_1-180}{6}
\sum_{j=-4}^{k_1}j^2 with k_1 \gt 0
We know that:
\sum_{j=1}^{n}j^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}
Since the 0^2 term adds nothing to the sum, it can be removed:
\sum_{j=-4}^{k_1}j^2=\sum_{j=1}^{k_1}j^2+ \sum_{j=-4}^{-1}j^2
Since j^2=\left(-j\right)^2 :
\sum_{j=-4}^{-1}j^2=\sum_{j=1}^{4}j^2
Therefore:
\sum_{j=-4}^{k_1}j^2=\sum_{j=1}^{k_1}j^2+ \sum_{j=1}^{4}j^2=\dfrac{k_1\cdot\left(k_1+1\right)\cdot\left(2k_1+1\right)}{6}+\dfrac{4\cdot5\cdot9}{6}
\sum_{j=-4}^{k_1}j^2=\dfrac{k_1\cdot\left(k_1+1\right)\cdot\left(2k_1+1\right)}{6}+30
\sum_{j=-4}^{k_1}j^2=\dfrac{2k_1^3+3k_1^2+k_1+180}{6}
\sum_{j=k_1}^{k_2}j^2 with 0 \lt k_1 \lt k_2
We know that:
\sum_{j=1}^{n}j^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}
Since the 0^2 term adds nothing to the sum, it can be removed:
\sum_{j=k_1}^{k_2}j^2=\sum_{j=1}^{k_2}j^2- \sum_{j=1}^{k_1}j^2
Therefore:
\sum_{j=k_1}^{k_2}j^2=\sum_{j=1}^{k_2}j^2- \sum_{j=1}^{k_1}j^2=\dfrac{k_2\cdot\left(k_2+1\right)\cdot\left(2k_2+1\right)}{6}-\dfrac{k_1\cdot\left(k_1+1\right)\cdot\left(2k_1+1\right)}{6}
\sum_{j=k_1}^{k_2}j^2=\dfrac{2\left(k_2^3-k_1^3\right)+3\left(k_2^2-k_1^2\right)+k_2-k_1}{6}