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# What is the fifth term in the binomial expansion of (x + 5)^{8}?

**Solution:**

The binomial expansion can be done using the formula

\((a+b)^{n}=\sum_{k=0}^{n} nC_{k}a^{n-k}b^{k}\)

In the expansion of (a + b)^{n} there are n + 1 terms

Sum of indices of a and b is equal to the n in every term of the expansion.

The general term T_{r + 1} of the binomial expansion (a + b)^{n} is \(T_{r+1}={n}C_{r}a^{n-r}b^{r}\)

We have to find the r value of the 5^{th} term

⇒ r + 1 = 5

⇒ r = 5 - 1

⇒ r = 4

So the 5th term of (x + 5)^{8} is T_{5} = T_{4+1}

\(\\={8}C_{4}.x^{8-4}.5^{4} \\ \\=\frac{8.7.6.5}{4.3.2.1}.x^{4}.625 \\ \\=625.70x^{4} \\ \\=43750x^{4}\)

Therefore, the fifth term is 43750x^{4}.

## What is the fifth term in the binomial expansion of (x + 5)^{8}?

**Summary:**

The fifth term in the binomial expansion of (x + 5)^{8} is 43750x^{4}.

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