@Ameya Daigavane
–
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.
–
Calvin Lin
Staff
·
1 year, 2 months ago

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TopNewest\[ S_n = \frac{1 - (1 - a)(1 - 2a)(1- 3a)\ldots(1-na)}{a} \] 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. – Ameya Daigavane · 1 year, 3 months ago

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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. – Calvin Lin Staff · 1 year, 2 months ago

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– Akhilesh Prasad · 1 year, 3 months ago

I was looking for more of an Algebraic proofLog in to reply

– Ameya Daigavane · 1 year, 3 months ago

I've added the motivation for the sum. Check it out.Log in to reply

– Akhilesh Prasad · 1 year, 3 months ago

Can you also see the other solution that i postedLog in to reply

– Akhilesh Prasad · 1 year, 3 months ago

Thanks a lot, thats what i needed.Log in to reply

@Rishabh Cool, @Svatejas Shivakumar, @Siddhartha Srivastava, @Ameya Daigavane – Akhilesh Prasad · 1 year, 3 months ago

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