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=(8!)/(4!4!)\cdot x^4 \cdot5^4\\ T_5=(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

`T_5=43750x^(4)`

Explanation:

Given:
`(x+5)^8`

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

Formula:

If
`(a+b)^n` then

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=(8!)/(4!\cdot 4!)\cdot x^(4)\cdot 625`
`\because ^nC_r=(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