Asked by Julie
In the binomial series from (a+b)^5, the powers of a decrease by 1 each term, and the powers of b increase by 1. If you carry on this pattern beyond te "end" of the series you get
a^5 + _a^4 b + _a^3 b^2 + _a^2 b^3 + _a^1 b^4 + _a^0 b^5 + _a^-1 b^6 + _a^-2 b^7 +...
where the spaces are for the coefficients.
a. What will be the coefficients for terms beyond 8? How do you know?
a^5 + _a^4 b + _a^3 b^2 + _a^2 b^3 + _a^1 b^4 + _a^0 b^5 + _a^-1 b^6 + _a^-2 b^7 +...
where the spaces are for the coefficients.
a. What will be the coefficients for terms beyond 8? How do you know?
Answers
Answered by
Reiny
First of all there would only be 6 terms in the expansion using the exponent of 5
Are you familiar with Pascal's Triangle?
If not, google it, and you will find that you need the coefficients of
1 5 10 10 5 1
that is:
a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5a b^4 + b^5
these coefficients can also be written in the form
C(5,0) C(5,1) C(5,2) C(5,3) C(5,4) and C(5,5)
where C(n,r) is defined as n!/(r!(n-r)!)
since n! is only defined for whole numbers of n
expressions such as C(5,6) are undefined and don't exist, so only 6 terms would exist for your expansion.
Are you familiar with Pascal's Triangle?
If not, google it, and you will find that you need the coefficients of
1 5 10 10 5 1
that is:
a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5a b^4 + b^5
these coefficients can also be written in the form
C(5,0) C(5,1) C(5,2) C(5,3) C(5,4) and C(5,5)
where C(n,r) is defined as n!/(r!(n-r)!)
since n! is only defined for whole numbers of n
expressions such as C(5,6) are undefined and don't exist, so only 6 terms would exist for your expansion.
There are no AI answers yet. The ability to request AI answers is coming soon!