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)+......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 nn terms.

#Algebra

Note by Akhilesh Prasad
4 years, 11 months ago

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Comments

Sn=1−(1−a)(1−2a)(1−3a)…(1−na)a S_n = \frac{1 - (1 - a)(1 - 2a)(1- 3a)\ldots(1-na)}{a} Use induction to prove easily as, Sk+(k+1)(1−a)(1−2a)(1−3a)…(1−ka)=1−(1−a)(1−2a)(1−3a)…(1−ka)(1−a(k+1))a=Sk+1 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: tk=k(1−a)(1−2a)(1−3a)…(1−(k−1)a) t_k = k(1 - a)(1 - 2a)(1- 3a)\ldots(1-(k - 1)a)

Write k k as the following:

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

so,

tk=(1−a)(1−2a)(1−3a)…(1−(k−1)a)a−(1−a)(1−2a)(1−3a)…(1−ka)a 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,

Sn=1−(1−a)(1−2a)(1−3a)…(1−na)a S_n = \frac{1 - (1 - a)(1 - 2a)(1- 3a)\ldots(1-na)}{a}

as claimed.

Ameya Daigavane - 4 years, 11 months ago

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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 Sk+1−Sk=tkS_{k+1}-S_k = t_k , which follows easily from the factorization.

Calvin Lin Staff - 4 years, 11 months ago

I was looking for more of an Algebraic proof

Akhilesh Prasad - 4 years, 11 months ago

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

Ameya Daigavane - 4 years, 11 months ago

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@Ameya Daigavane – Thanks a lot, thats what i needed.

Akhilesh Prasad - 4 years, 11 months ago

@Ameya Daigavane – Can you also see the other solution that i posted

Akhilesh Prasad - 4 years, 11 months ago
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