IIT JEE , AIEEE Forum - Mathematics , Algebra

man111

how can i solve these type of question.

(1) 1⁶+2⁶+3⁶+4⁶............n⁶ = An⁷+Bn⁶+Cn⁵+Dn⁴+En³+Fn²+Gn+E

what will be the value of A,B,C,D,E,F

konichiwa2x can you solve this question.

and gave a general formula for this question means

(1) 1^r+2^r+3^r+4^r............n^r = A n^(r+1)+Bn^r......................

konichiwa2x

I am flattered that you ask for me. I will give the question a try.

By the way, the last part of the question..is it Fn^2+Gn+H or Fn^2+Gn+E?

man111

It is in decreasing power of x matlab 7,6,5,4,3,2,1

and A,B,C,D,E,F,G are cofficient

Iam asking bcz u r so amazing solve all of my question.

konichiwa2x

Are the answers

,

?

konichiwa2x

To solve this question you need knowldge of bernoulli numbers B_n which were first discovered in connection with the closed forms of the sums

$sum_{k=0}^{m-1} k^n = 0^n + 1^n + 2^n + cdots + {(m-1)}^n$

for various fixed values of n. The closed forms are always polynomials in m of degree n + 1. The coefficients of these polynomials are closely related to the Bernoulli numbers, as follows:

$sum_{k=0}^{m-1} k^n = {1over{n+1}}sum_{k=0}^n{n+1choose{k}} B_k m^{n+1-k}.$

Here

.

For example, taking n to be 1, we have 0 + 1 + 2 + ... + (m ? 1) = (1/2) (B₀ m² + 2 B₁ m¹) = 1/2 (m² ? m).

Bernoulli numbers may be calculated by using the following recursive formula:

$sum_{j=0}^m{m+1choose{j}}B_j = 0$

for m > 0, and B₀ = 1..

For more info, check out this link:

http://en.wikipedia.org/wiki/Bernoulli_numbers

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In this question,

Where the bernoulli numbers

,

.

Now, a smart thing to do would be to take the summation only till

. Otherwise, simplification becomes difficult with terms like

on the R.H.S.

(on putting the values of the coefficients)

Now add

to both side to get,

vineetnegi

the bernouli number stuff is called faulkhabers series

but i dont think that we need to use anything that advanced

this the question of undertermined coefficient

this is an identity

EDIT :see the process below

konichiwa2x

It becomes very tedious with substitutions. I personally am not fond of evaluating 3^7 or so on..anyway 3 equations are not sufficient.

konichiwa2x

For the 2nd part of your question, i.e the general case:

Hence coefficient of

The coefficients give you the value of A,C,D,E.... and so on.

Now add

to both sides of the equation.
Hence, the coefficient of

will be

. This is the value of B.

vineetnegi

true that three equation stuff is too complicated!!!!!!!!!

and lengthy too!!!!!!!!!!!!!!!

using principle of undertermined coeficient

1^6.......+n^6=An^7+Bn^6...................+H

right

1^6+............+n^6+(n+1)^6=A(n+1)^7............+H

2eq-1eq

(n+1)^7=A((n+1)^7-n^7)..................+G((n+!)-n)

expand and group

(6C0+6C1n+........)=A(7C6)n^6+(A7C5+B6C5)n^5+.............

equating coefinient of like terms on both side

a=6C6/7C6

A=1/7

proceed same way to for all

by the way given three equation in any no. of variable

provided thier degree is same and they are non consistent in all variable (which is true inthis case)

can be solved

konichiwa2x

ya, 3 equations are enough since they are not consistent. missed that point earlier.

By the way man111, are the answers correct?

man111

yes absolutely correct.

and for second i thing it is correct.

and i am just asking for a general formula.

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