Verified Answer :

konrad509

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

isyllus

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

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