Sigma Notation

4³ + 5³ + 6³ + ... + 19³

This is a series.
The cubed numbers go from 4 to 19.

So write
Σ,
k = 4 at the bottom of the sigma,
and 19 at the top of the sigma.

See the terms.
4³, 5³, 6³, ..., 19³

Then a_k = k³.

So write a^k: k³.

Sequence

So this is the answer.

To write the series in sigma notation,
find a_k.

See the terms of the series:
1, 3, 5, 7, ... .

The first term, a, is 1.

1 + 2 = 3
3 + 2 = 5
5 + 2 = 7

So write, the d, +2
between the terms.

This is an arithmetic sequence.

a = 1
d = 2

So a_k = 1 + (k - 1)⋅2.

Arithmetic Sequence

+(k - 1)⋅2 = +2k - 2

1 + 2k - 2 = 2k - 1

So a_k = 2k - 1.

See the terms of the given series.

There are 7 terms.

So k goes from 1 to 7:
(given) = a₁ + a₂ + ... + a₇.

a_k = 2k - 1
k goes from 1 to 7.

Then the given series is
Σ
k goes from 1 to 7
2k - 1.

So this is the answer.

(k + 1) - k = 1: constant.

So
1/[k(k + 1)] = [1/[(k + 1) - k]]⋅[1/k - 1/(k + 1)].

Partial Fraction Decomposition: Formula

1/[(k + 1) - k] = 1/1

[1/1]⋅[1/k - 1/(k + 1)] = [1/k - 1/(k + 1)]

Expand the given series.

k = 1

(1/1 - 1/2)

k = 2

+(1/2 - 1/3)

k = 3

+(1/3 - 1/4)

Write ... .

k = 99

+(1/99 - 1/100)

So the given series is
(1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/99 - 1/100).

Cancel -1/2 and 1/2.
Cancel -1/3 and 1/3.
Cancel -1/4 and 1/4.
...
And cancel -1/99 and 1/99.

Then 1/1 and -1/100 remain.
So write 1/1 - 1/100.

When cancelling like this,
the remaining terms are located symmetrically.
(In this case,
the first and the last terms remain.)

1/1 = 100/100

100/100 - 1/100 = 99/100

So 99/100 is the answer.