Question

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Answer 1

Given expression is (a+b)^8 and we need to find the 5th term of this series.

Explanation -

The general binomial expression is given as

`(x+y)^n=^nC_0x^ny^0+^nC_1x^(n-1)y^1+^nC_2x^(n-2)y^2+............upto\text{ x}^0`

So the given expression can be written as

`(a+b)^8=^8C_0a^8b^0+^8C_1a^7y^1+^8C_2a^6b^2+^8C_3a^5b^3+^8C_4a^4b^4+...`

So the 5th term will be

`\begin{gathered} 5th\text{ term = }^8C_4a^4b^4 \\ The\text{ general comb}\imaginaryI\text{nat}\imaginaryI\text{on formula }\imaginaryI\text{s = }^nC_r=(n!)/((n-r)!r!) \end{gathered}`

Then we have

`\begin{gathered} 5th\text{ term = }(8!)/((8-4)!4!)a^4b^4 \\ 5th\text{ term = }(8*7*6*5)/(4!)a^4b^4 \\ 5th\text{ term = }(8*7*6*5)/(4*3*2*1)a^4b^4=2*7*5* a^4b^4=70a^4b^4=70(ab)^4 \end{gathered}`So the final answer is 70(ab)^4