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NARA [144]
3 years ago
9

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

Mathematics
2 answers:
AlekseyPX3 years ago
5 0
(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

Artyom0805 [142]3 years ago
4 0

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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