Binomial Theorem: Find 6th term of (a - 2b)^8

[kpx001]

How would i find the sixth term of (a-2b)^8 ? i tried two different formulas but i get different results.

(n,r) = (a^n-k)(b^k) (8,5)= (a^8-5)(-2b)^5 = -32a^3b^5

and

C(n,r) = n!/(r!(n-r)!) C(8.5) = 8! / (5!(8-5)!)

i get 56 and i dunno what to do with that number.

 

[galactus]

Use the binomial expansion

\(\displaystyle \L\\(a+b)^{n}=a^{n}+\left(\begin{array}{c}n\\1\end{array}\right)a^{n-1}b+\left(\begin{array}{c}n\\2\end{array}\right)a^{n-2}b^{2}+......................+\left(\begin{array}{c}n\\k\end{array}\right)a^{n-k}b^{k}+..........\\+\left(\begin{array}{c}n\\k+1\end{array}\right)ab^{n-1}+b^{n}\)

b=-2b in this case:

\(\displaystyle \L\\a^{8}-16a^{7}b+112a^{6}b^{2}-448a^{5}b^{3}+.....\)

See the pattern?. What are you calling the 6th term.

The coefficient of \(\displaystyle a^{3}b^{5}\)?. That's the 6th term from the left.

 

[stapel]

i get 56 and i dunno what to do with that number.

How are you getting "56"? For which part of your answer (or is that your entire answer)? :?:

Note: It's a lot easier for tutors to help you find any errors and to provide you with advice if you show what you have done, rather than just making reference to some intermediate or final result. :wink:

Thank you!

Eliz.

 

[pka]

It is important to note that there is no universal agreement on the ordering of the terms in any expansion. The sixth term for someone may not be what another would call the sixth term.