×

# Help:Sequence and Series

Find the sum of the series: $$1 + 2(1-a) + 3(1-a)(1-2a) + 4(1-a)(1-2a)(1-3a) + 5(1-a)(1-2a)(1-3a)(1-4a)+......$$ to $$n$$ terms.

1 year, 5 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

$S_n = \frac{1 - (1 - a)(1 - 2a)(1- 3a)\ldots(1-na)}{a}$ Use induction to prove easily as, $S_k + (k + 1)(1 - a)(1 - 2a)(1- 3a)\ldots(1-ka) = \frac{1 - (1 - a)(1 - 2a)(1- 3a)\ldots(1-ka)(1 - a(k + 1))}{a} = S_{k + 1}$

EDIT: $t_k = k(1 - a)(1 - 2a)(1- 3a)\ldots(1-(k - 1)a)$

Write $$k$$ as the following:

$$k = \dfrac{1 - (1 - ka)}{a} = \dfrac{1}{a} - \dfrac{(1 - ka)}{a}$$

so,

$t_k = \dfrac{(1 - a)(1 - 2a)(1- 3a)\ldots(1-(k - 1)a)}{a} - \dfrac{(1 - a)(1 - 2a)(1- 3a)\ldots(1 - ka)}{a}$

which when summed, telescopes to,

$S_n = \frac{1 - (1 - a)(1 - 2a)(1- 3a)\ldots(1-na)}{a}$

as claimed.

- 1 year, 5 months ago

I was initially amazed that the answer was so nice, certainly wasn't expecting that.

A slightly better way to present it would be to show that $$S_{k+1}-S_k = t_k$$, which follows easily from the factorization.

Staff - 1 year, 5 months ago

I was looking for more of an Algebraic proof

- 1 year, 5 months ago

I've added the motivation for the sum. Check it out.

- 1 year, 5 months ago

Can you also see the other solution that i posted

- 1 year, 5 months ago

Thanks a lot, thats what i needed.

- 1 year, 5 months ago