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Finite Sum
Related Topics
Wize University Calculus 1 Textbook > Integrals
Finite Sums
4 Activities
Evaluate the following sum:
∑
i
=
10
65
k
(
k
−
1
)
\displaystyle \sum_{i=10}^{65}k(k-1)
i
=
10
∑
65
k
(
k
−
1
)
Answer
I don't know
Check Submission
More Finite Sums Questions:
Finite Sums
Evaluate
∑
i
=
5
10
[
2
i
−
3
]
\displaystyle\sum_{i=5}^{10}\left[2i-3\right]
i
=
5
∑
10
[
2
i
−
3
]
.
Finite Sums: Sigma Notation
∑
i
=
0
30
sin
(
i
π
2
)
\sum_{i=0}^{30}\sin\left(\frac{i\pi}{2}\right)
∑
i
=
0
30
sin
(
2
iπ
)
equals
Practice: Telescoping Sum (~F2017 Final Q22)
Practice: Sigma Notation
Evaluate
Σ
i
=
1
2000
(
1
i
+
1
−
1
i
+
2
)
\Sigma_{i=1}^{2000}\ \left(\frac{1}{i+1}-\frac{1}{i+2}\right)
Σ
i
=
1
2000
(
i
+
1
1
−
i
+
2
1
)
Practice: Sigma Notation (~F2017 Final Q21)
Practice: Sigma Notation
Evaluate
Σ
i
=
10
20
(
3
i
−
4
)
\Sigma_{i=10}^{20}\ \left(3i-4\right)
Σ
i
=
10
20
(
3
i
−
4
)
.
Finite Sum with Trig
Evaluate the following sum:
∑
i
=
2
5
sin
(
i
π
2
)
\displaystyle \sum_{i=2}^{5}\sin{\left(i\frac{\pi}{2}\right)}
i
=
2
∑
5
sin
(
i
2
π
)
Finite Sum with Trig
Evaluate the sum:
∑
j
=
1
20
j
sin
(
j
π
2
)
\displaystyle \sum_{j=1}^{20}j\sin{\left(j\frac{\pi}{2}\right)}
j
=
1
∑
20
j
sin
(
j
2
π
)
Finite Sums
Evaluate
∑
i
=
2
100
1
i
−
1
−
1
i
\sum_{i=2}^{100}\ \frac{1}{i-1}-\frac{1}{i}
∑
i
=
2
100
i
−
1
1
−
i
1
.
Finite Sums
Evaluate
∑
i
=
5
10
[
2
i
−
3
]
\displaystyle\sum_{i=5}^{10}\left[2i-3\right]
i
=
5
∑
10
[
2
i
−
3
]
.
Evaluate
3
⋅
Σ
i
=
5
10
[
(
2
i
)
2
−
2
(
i
−
1
)
]
3\cdot\Sigma_{i=5}^{10}\ \left[\left(2i\right)^2-2\left(i-1\right)\right]
3
⋅
Σ
i
=
5
10
[
(
2
i
)
2
−
2
(
i
−
1
)
]
Evaluate
Σ
i
=
1
100
(
1
3
i
−
1
−
1
3
i
)
\Sigma_{i=1}^{100}\ \left(\frac{1}{3^{i-1}}-\frac{1}{3^i}\right)
Σ
i
=
1
100
(
3
i
−
1
1
−
3
i
1
)
.
Finite Sums
Evaluate
∑
i
=
1
20
(
1
+
4
i
)
\sum_{i=1}^{20}(1+4i)
∑
i
=
1
20
(
1
+
4
i
)
The sum of the series
∑
n
=
1
N
(
3
+
∑
i
=
1
n
i
)
=
\displaystyle \sum_{n=1}^{N} {\left(3+\sum_{i=1}^n{i}\right)}=
n
=
1
∑
N
(
3
+
i
=
1
∑
n
i
)
=
Practice: Telescoping Sums
Evaluate
∑
i
=
1
2000
(
1
i
+
1
−
1
i
+
2
)
\displaystyle \sum_{i=1}^{2000}\ \left(\frac{1}{i+1}-\frac{1}{i+2}\right)
i
=
1
∑
2000
(
i
+
1
1
−
i
+
2
1
)
Practice: Sigma w/ Limits
Evaluate
lim
n
→
∞
∑
i
=
1
n
(
3
n
4
)
⋅
[
i
(
2
i
2
−
1
)
+
1
]
\displaystyle \lim_{n\rightarrow\infty}\ \sum_{i=1}^n\left(\frac{3}{n^4}\right)\cdot\left[i\left(2i^2-1\right)+1\right]
n
→
∞
lim
i
=
1
∑
n
(
n
4
3
)
⋅
[
i
(
2
i
2
−
1
)
+
1
]
Practice: Sigma Notation
Evaluate
∑
i
=
10
20
(
3
i
−
4
)
\displaystyle \sum_{i=10}^{20}\ \left(3i-4\right)
i
=
10
∑
20
(
3
i
−
4
)
.
Finite Sum with Trig
Evaluate the following sum:
∑
i
=
2
5
sin
(
i
π
2
)
\displaystyle \sum_{i=2}^{5}\sin{\left(i\frac{\pi}{2}\right)}
i
=
2
∑
5
sin
(
i
2
π
)
Finite Sum
Evaluate the following sum:
∑
i
=
10
65
k
(
k
−
1
)
\displaystyle \sum_{i=10}^{65}k(k-1)
i
=
10
∑
65
k
(
k
−
1
)
Finite Sum with Trig
Evaluate the sum:
∑
j
=
1
20
j
sin
(
j
π
2
)
\displaystyle \sum_{j=1}^{20}j\sin{\left(j\frac{\pi}{2}\right)}
j
=
1
∑
20
j
sin
(
j
2
π
)
Evaluate
Σ
i
=
1
100
(
1
3
i
−
1
−
1
3
i
)
\Sigma_{i=1}^{100}\ \left(\frac{1}{3^{i-1}}-\frac{1}{3^i}\right)
Σ
i
=
1
100
(
3
i
−
1
1
−
3
i
1
)
.
Evaluate
3
⋅
Σ
i
=
5
10
[
(
2
i
)
2
−
2
(
i
−
1
)
]
3\cdot\Sigma_{i=5}^{10}\ \left[\left(2i\right)^2-2\left(i-1\right)\right]
3
⋅
Σ
i
=
5
10
[
(
2
i
)
2
−
2
(
i
−
1
)
]
Finite Sums
Evaluate
Σ
i
=
3
10
[
i
(
2
+
i
)
−
1
]
\Sigma_{i=3}^{10}\ \left[i\left(2+i\right)-1\right]
Σ
i
=
3
10
[
i
(
2
+
i
)
−
1
]
Finite Sums
Evaluate
Σ
i
=
1
1000
1
e
i
−
1
e
i
−
1
\Sigma_{i=1}^{1000}\ \frac{1}{e^i}-\frac{1}{e^{i-1}}
Σ
i
=
1
1000
e
i
1
−
e
i
−
1
1
.
Finite Sums: Sigma Notation
∑
i
=
0
30
sin
(
i
π
2
)
\sum_{i=0}^{30}\sin\left(\frac{i\pi}{2}\right)
∑
i
=
0
30
sin
(
2
iπ
)
equals
Practice: Sigma Notation (~F2017 Final Q21)
Practice: Sigma Notation
Evaluate
Σ
i
=
10
20
(
3
i
−
4
)
\Sigma_{i=10}^{20}\ \left(3i-4\right)
Σ
i
=
10
20
(
3
i
−
4
)
.
Practice: Telescoping Sum (~F2017 Final Q22)
Practice: Sigma Notation
Evaluate
Σ
i
=
1
2000
(
1
i
+
1
−
1
i
+
2
)
\Sigma_{i=1}^{2000}\ \left(\frac{1}{i+1}-\frac{1}{i+2}\right)
Σ
i
=
1
2000
(
i
+
1
1
−
i
+
2
1
)