How do you find the 2nd term in the expansion of ${(y - 2 x)}^{4}$ $(y-2x)^4$ ?

Question

14424 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-01T18:03:29

$The reqd. term = - 8 y^{3} x$ $"The reqd. term"=-8y^3x$ .

In the Binomial Expansion of ${(a + b)}^{n}$ $(a+b)^n$ , the general ${(r + 1)}^{t h}$ $(r+1)^(th)$ term,

i.e., $T_{r + 1}$ $T_(r+1)$ , is given by,

$T_{r + 1} =_{n} C_{r} \cdot a^{n - r} \cdot b^{r}$ $T_(r+1)=""_nC_r*a^(n-r)*b^r$ .

In our example, $n = 4, a = y, b = - 2 x, r + 1 = 2, i . e ., r = 1$ $n=4, a=y, b=-2x, r+1=2, i.e., r=1$

$:. T_(1+1)=T_2=""_4C_1*y^(4-1)*(-2x)^1$

$=4*y^3*(-2x)$

$=-8y^3x$ .

How do you find the 2nd term in the expansion of (y-2x)^4(y−2x)4(y-2x)^4?