Question

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

Answer 1

$(a+b)^n\\ \displaystyle T_r=\binom{n}{r-1}a^{n-r+1}b^{r-1}\\\\ T_5=\binom{8}{5-1}x^{8-5+1}\cdot5^{5-1}\\ T_5=\binom{8}{4}x^4\cdot5^4\\ T_5=\dfrac{8!}{4!4!}\cdot x^4 \cdot5^4\\ T_5=\dfrac{5\cdot6\cdot7\cdot8}{2\cdot3\cdot4}\cdot x^4\cdot5^4\\ T_6=2\cdot7\cdot x^4\cdot5^5\\ T_6=43750x^4$

Answer 2

Answer:

$T_5=43750x^{4}$    

Step-by-step explanation:

Given: $(x+5)^8$

It is binomial expansion. It has two term with power 8.

Formula:

If $(a+b)^n$ then $T_{r+1}=^nC_ra^^{n-r}b^r$

This is general formula of the expansion.

For fifth term, T₅

$T_5=T_{r+1}$

So, r=4

$(x+5)^8$

Put r=4 and n=8, a=x, b=5 into formula and we get

$T_5=^8C_4\cdot x^{8-4}\cdot 5^4$

$T_5=\dfrac{8!}{4!\cdot 4!}\cdot x^{4}\cdot 625$            $\because ^nC_r=\dfrac{n!}{r!(n-r)!}$

$T_5=70\cdot x^{4}\cdot 625$

$T_5=43750x^{4}$    

Hence, The fifth term of the given binomial is 43750x⁴