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# 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 ago

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$S_n = \frac{1 - (1 - a)(1 - 2a)(1- 3a)\ldots(1-na)}{a}$ 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 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 · 12 months ago

I was looking for more of an Algebraic proof · 1 year ago

I've added the motivation for the sum. Check it out. · 1 year ago

Can you also see the other solution that i posted · 1 year ago

Thanks a lot, thats what i needed. · 1 year ago