Could you please tell me how to find the summation of a series whose difference of difference (2nd order difference) of successive terms is a constant.

Ex- 3+4+9+18+31+...... 1st order difference is 4-3=1

9-4=5

18-9=9. 31-18=13 Second order difference is 5-1=4, 9-5=4, 13-9=4

Hence second order difference is a constant.so how do you find the summation of such a series upto n terms.

Also it would be helpful if you could explain a general procedure if the nth order difference was a constant.

Thanks in advance.

P.s would be grateful to anyone who would reply at the earliest as I need it for an exam the day after tomorrow

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TopNewestI think the \(n\)th term of these sequences (constant second order difference) can be expressed in the form \(an^2+bn+c\), you can then find \(a, b, c\) by substituting in values of n: For example with your example \(a+b+c=3, 4a+2b+c=4, 9a+3b+c=9\), solving yields \(a=2, b=-5, c=6\), so the nth term is \(2n^2-5n+6\), you can then use the sum of squares formula \(\frac{n(n + 1)(2n + 1)}{6}\) and the sum of first \(n\) positive integers formula \(\frac{n(n+1)}{2}\) to evaluate the sum.

I suspect (but currently have no proof) that if the \(k\)th order difference is constant, then each member of the series is a polynomial of order \(k\), so you could get the sum by using formulae for the sum of the \(k\)th, \(k-1\)th...\(1\)st powers of the first \(n\) positive integers. – Clifford Wilmot · 4 years, 1 month ago

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– Paramjit Singh · 4 years, 1 month ago

Definitely, your suspicion is correct since for a \(k\)th order polynomial you can create a finite difference table whose \(k\)th row is constant...Log in to reply

– Pranav Chakravarthy · 4 years, 1 month ago

Thanks.but could you elaborate as to how you deduced that it must of the form an^2+bn+c.Log in to reply

– Sebastian Garrido · 4 years, 1 month ago

you can think of it as a function whose second derivative is constant, hence the polynomial for the sequence will be a quadratic.Log in to reply

– Clifford Wilmot · 4 years, 1 month ago

This is how I originally thought about it.Log in to reply

SEE..here we see there are two series..one is the one that you gave..another is the series of the differences..let us call the series you gave {a}n..(subscript n) and the series of the differences 1,5,9,13,.. as {b}n...now if you notice

{a}n - {a}n-1 = {b}n-1

i.e. the difference of the nth term and the n-1 th term of the series in the question is equal to the n-1 th term of the series of the differences (1,3,5,.. i.e. {b}n)

therefore,

[summing them all up] {a}n - {a}1 = {b}n-1 + {b}n-2 + ..... + {b}1 .....................(eqn 1)_________now {b}n is a general arithmetic progression series..we can find the general term of {b}n {b}n = 4n-3 find summation of {b}n, where n varies from n-1 to 1

so we get to know the right hand side of equation 1 {a}1=3 [given] therefore find {a}n

it'll be {a}n=2n^2 - 5n + 6 – Preetam Sengupta · 4 years, 1 month ago

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– Pranav Chakravarthy · 4 years, 1 month ago

Thanks a lot that was very neatly explained.Log in to reply

pranav, the formula for the summation of this series is, n.a +(∑(n-1))a '+(∑(n-2))d where, a is the first number of the series a' is the first number of the secondary series which is obtained by 1st order difference d is the constant which is obtained by the 2nd order difference a is 3 , a' is 1 , d is 4 in your given example note this formula is only applicable where n is greater than 1 this is my 1st post, hope it is ueful for u – Saichandra Rockzz · 4 years, 1 month ago

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no -_- – The Destroyer · 4 years, 1 month ago

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how to simple continued fraction of 41/29 – Avijit Pandey · 4 years, 1 month ago

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