How do you find the 6th term of ${(a + b)}^{8}$ $(a + b)^8$ ?

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Answer 1 · 2016-03-19T19:54:37

You must use the formula $t_{r + 1} = n C_{r} \times a^{n - r} \times b^{r}$ $t_(r + 1) = nC_r xx a^(n - r) xx b^r$

n is the expression's exponent

$r + 1 = 6$ $r + 1 = 6$

$r = 5$ $r = 5$

$n C_{r}$ $nC_r$ is calculated by the formula $\frac{n!}{(n - r)! (r!)}$ $(n!)/((n - r)!(r!))$

n = 8 and r = 5. Calculating, we get:

$\frac{8!}{(3! \times 5!)}$ $(8!)/((3! xx 5!)$

= 56

$t_{6} = 56 \times a^{8 - 5} \times b^{5}$ $t_6 =56 xx a^(8 - 5) xx b^5$

$t_{6} = 56 a^{3} b^{5}$ $t_6 = 56a^3b^5$

Hopefully this helps!

How do you find the 6th term of (a+b)8(a + b)^8 ?