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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

Note by Pranav Chakravarthy
3 years, 11 months ago

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I 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 · 3 years, 11 months ago

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@Clifford Wilmot 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... Paramjit Singh · 3 years, 11 months ago

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@Clifford Wilmot Thanks.but could you elaborate as to how you deduced that it must of the form an^2+bn+c. Pranav Chakravarthy · 3 years, 11 months ago

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@Pranav Chakravarthy you can think of it as a function whose second derivative is constant, hence the polynomial for the sequence will be a quadratic. Sebastian Garrido · 3 years, 11 months ago

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@Sebastian Garrido This is how I originally thought about it. Clifford Wilmot · 3 years, 11 months ago

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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,

         {a}n   -  {a}n-1 = {b}n-1
         {a}n-1 - {a}n-2 = {b}n-2
         .....
         .......
         .....
         {a}2    - {a}1    = {b}1

_________ [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 · 3 years, 11 months ago

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@Preetam Sengupta Thanks a lot that was very neatly explained. Pranav Chakravarthy · 3 years, 11 months ago

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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 · 3 years, 11 months ago

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no -_- The Destroyer · 3 years, 11 months ago

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how to simple continued fraction of 41/29 Avijit Pandey · 3 years, 11 months ago

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