What is the 5th term of the expansion of $(2x+7)^8$ using the Binomial Theorem?

[karush]

Find the 5th term of $(2x+7)^8$

Assume Binomial Theorem can be used on this.
not sure what determines the 5th term

 

[MarkFL]

Find the 5th term of $(2x+7)^8$

Assume Binomial Theorem can be used on this.
not sure what determines the 5th term

Yes, the binomial theorem is your friend here...and it states:

\(\displaystyle (a+b)^n=\sum_{k=0}^n\left({n \choose k}a^{n-k}b^k\right)\)

And so the $m$th term of the expansion would be for $m=k+1$...can you proceed?

 

[karush]

${5}^{th}$ term =
$$2689120{x}^{4}$$

 

[MarkFL]

I get:

\(\displaystyle {8 \choose 4}(2x)^{8-4}(7)^4=70\cdot16x^4\cdot7^4=2689120x^4\checkmark\)

 

[suluclac]

If this was on the exam, some students could multiply 2x + 7 itself 8 times and get the desired result.
Hopefully, the student has studied the binomial theorem before the exam!