What is the fourth term in the expansion of ${(2 x - y)}^{5}$ $(2x-y)^5$ ?

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What is the fourth term in the expansion of ${(2 x - y)}^{5}$ $(2x-y)^5$ ?

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Answer 1 · 2016-02-13T05:27:08

You can find this by using the formula $t_{r + 1} =_{n} C_{r} \times a^{n - r} \times b^{r}$ $t_(r + 1) =_nC_r xx a^(n - r) xx b^r$

Explanation:

r can be found by solving the simple equation $t_{r + 1} = 4$ $t_(r + 1) = 4$ We can immediately eliminate the t, and we find that r is 3. n is the exponent, which is 5 in this case. Note the n and the r are in base at the C.

$t_{4} =_{5} C_{3} \times 2 x^{5 - 3} \times - y^{3}$ $t_4 =_5C_3 xx 2x^(5 - 3) xx -y^3$

$t_{4} = 10 \times 4 x^{2} (- y^{3})$ $t_4 = 10 xx 4x^2(-y^3)$

$t_{4} = - 40 x^{2} y^{3}$ $t_4 = -40x^2y^3$

Your fourth term is $- 40 x^{2} y^{3}$ $-40x^2y^3$ .

Practice exercises:

Find the 7th term in ${(- 2 x + 3 y)}^{12}$ $(-2x + 3y)^12$
Find the middle term of ${(3 x - y^{3})}^{8}$ $(3x - y^3)^8$

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What is the fourth term in the expansion of ${(2 x - y)}^{5}$ $(2x-y)^5$ ?

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What is the fourth term in the expansion of ${(2 x - y)}^{5}$ $(2x-y)^5$ ?