How do you find the 10th term of ${(x + 3)}^{12}$ $(x+3)^12$ ?

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How do you find the 10th term of ${(x + 3)}^{12}$ $(x+3)^12$ ?

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Answer 1 · 2016-11-03T01:33:14

The tenth term is $4, 330, 260 x^{3}$ $4,330,260x^3$ .

Explanation:

The nth term in the expansion of ${(a + b)}^{m}$ $(a + b)^m$ is given by $t_{k + 1} = 2_{n} C_{k} \times a^{n - k} \times b^{k}$ $t_(k+ 1)= color(white)(2)_nC_k xx a^(n - k) xx b^k$ .

We want to find the 10th term, but we need to find the value of $k$ $k$ to calculate it. Hence,

$k + 1 = 10$ $k + 1 = 10$

$k = 9$ $k = 9$

$t_{10} = 2_{12} C_{9} \times x^{12 - 9} \times 3^{9}$ $t_10 = color(white)(2)_12C_9 xx x^(12 - 9) xx 3^9$

$t_{10} = 220 x^{3} (19, 683)$ $t_10 = 220x^3(19,683)$

$t_{10} = 4, 330, 260 x^{3}$ $t_10 = 4,330,260x^3$

Hopefully this helps!

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How do you find the 10th term of ${(x + 3)}^{12}$ $(x+3)^12$ ?

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How do you find the 10th term of ${(x + 3)}^{12}$ $(x+3)^12$ ?