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

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The 5th term of the expansion of $(2x+7)^8$ using the Binomial Theorem is $2689120x^4$. This result is derived from the formula for the binomial expansion, which states that the $m$th term is given by ${n \choose k}a^{n-k}b^k$. Specifically, for this case, the calculation involves ${8 \choose 4}(2x)^{4}(7)^4$, resulting in the final term of $2689120x^4$. Understanding the application of the Binomial Theorem is crucial for accurately determining specific terms in polynomial expansions.

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

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

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

And so the $m$th term of the expansion would be for $m=k+1$...can you proceed?
 
${5}^{th}$ term =
$$2689120{x}^{4}$$

😍😍😍
 
I get:

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

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