What is the value of the following sums?
\sum_{i=4}^{11}i
We have the following formula:
\sum_{i=1}^{n}i=\dfrac{n\left(n+1\right)}{2}
We notice that:
\sum_{i=4}^{11}i=\sum_{i=1}^{11}i-\sum_{i=1}^{3}i
Therefore:
\sum_{i=4}^{11}i=\sum_{i=1}^{11}i-\sum_{i=1}^{3}i=\dfrac{11\left(12\right)}{2}-\dfrac{3\left(4\right)}{2}=66-6=60
\sum_{i=4}^{11}i=60
\sum_{i=10}^{50}i
We have the following formula:
\sum_{i=1}^{n}i=\dfrac{n\left(n+1\right)}{2}
We notice that:
\sum_{i=10}^{50}i=\sum_{i=1}^{50}i-\sum_{i=1}^{9}i
Therefore:
\sum_{i=10}^{50}i=\sum_{i=1}^{50}i-\sum_{i=1}^{9}i=\dfrac{50\left(51\right)}{2}-\dfrac{9\left(10\right)}{2}=1\ 275-45=1\ 230
\sum_{i=10}^{50}i=1\ 230
\sum_{i=5}^{150}i
We have the following formula:
\sum_{i=1}^{n}i=\dfrac{n\left(n+1\right)}{2}
We notice that:
\sum_{i=5}^{150}i=\sum_{i=1}^{150}i-\sum_{i=1}^{4}i
Therefore:
\sum_{i=5}^{150}i=\sum_{i=1}^{150}i-\sum_{i=1}^{4}i=\dfrac{150\left(151\right)}{2}-\dfrac{4\left(5\right)}{2}=11{,}325-10=11{,}315
\sum_{i=5}^{150}i=11{,}315
\sum_{i=-10}^{100}i
We have the following formula:
\sum_{i=1}^{n}i=\dfrac{n\left(n+1\right)}{2} where n \gt 0
We notice that:
\sum_{i=-10}^{100}i=\sum_{i=-10}^{0}i-\sum_{i=1}^{100}i
We can remove the i=0 term in the sum since adding zero does not affect the sum.
Additionally,
\sum_{i=-10}^{-1}i=-\sum_{i=1}^{10}i
Therefore:
\sum_{i=-10}^{100}i=\sum_{i=1}^{100}i+\sum_{i=-10}^{-1}i=\sum_{i=1}^{100}i-\sum_{i=1}^{10}i=\dfrac{100\left(101\right)}{2}-\dfrac{10\left(11\right)}{2}=5\ 050-55
\sum_{i=-10}^{100}i=4\ 995
\sum_{i=1}^{k_1}i
We have the following formula:
\sum_{i=1}^{n}i=\dfrac{n\left(n+1\right)}{2}
By substituting n=k_1 :
\sum_{i=1}^{k_1}i=\dfrac{k_1\left(k_1+1\right)}{2}
\sum_{i=5}^{k_1}i
We have the following formula:
\sum_{i=1}^{n}i=\dfrac{n\left(n+1\right)}{2}
We notice that:
\sum_{i=5}^{k_1}i=\sum_{i=1}^{k_1}i-\sum_{i=1}^{4}i
Therefore:
\sum_{i=5}^{k_1}i=\sum_{i=1}^{k_1}i-\sum_{i=1}^{4}i=\dfrac{k_1\left(k_1+1\right)}{2}-\dfrac{4\left(5\right)}{2}
\sum_{i=5}^{k_1}i=\dfrac{k_1\left(k_1+1\right)}{2}-10
\sum_{i=k_1}^{100}i, 0 \lt k_1 \lt 100
We have the following formula:
\sum_{i=1}^{n}i=\dfrac{n\left(n+1\right)}{2}
We notice that:
\sum_{i=k_1}^{100}i=\sum_{i=1}^{100}i-\sum_{i=1}^{k_1}i
Therefore:
\sum_{i=k_1}^{100}i=\sum_{i=1}^{100}i-\sum_{i=1}^{k_1}i=\dfrac{100\left(101\right)}{2}-\dfrac{k_1\left(k_1+1\right)}{2}
\sum_{i=k_1}^{100}i=5\ 050-\dfrac{k_1\left(k_1+1\right)}{2}