What is the 4th term in the expansion of (y+4x^3)^4?

(y+4x^3) ^4

2 Answers

256yx^9

Explanation:

It's (a+b)^n =[color(white)(o)^nC_0*a^(n-0)*b^0] + [color(white)(o)^nC_1*a^(n-1)*b^1] + [color(white)(o)^nC_2*a^(n-2)*b^2] +[color(white)(o)^nC_3*a^(n-3)*b^3] +.........

First term color(white)(ww) =>(color(white)(i)^4C_0)(y^4)(4x^3)^0
Second term color(white)(i) =>(color(white)(i)^4C_1)(y^3)(4x^3)^1
Third term color(white)(wu) =>(color(white)(i)^4C_2)(y^2)(4x^3)^2
Fourth term color(white)(ir) =>(color(white)(i)^4C_3)(y^1)(4x^3)^3

Fourth term =4*y*64x^9 color(white)(wwwwww) ["as " color(white)(i)^4C_3 = 4]

That is, 256yx^9

Apr 12, 2018

color(blue)(256yx^9)

Explanation:

For binomial expansions of the form (y+x)^n, with y descending, we have:

sum_(r=0)^(n)((n),(r))(y^(n-r)x^r)

Where:

((n),(r))=color(white)(0)^nC_(r)=(n!)/((r!(n-r)!)

We require the 4th term. The power of y in the 4th term will be 1:

So:

4-r=1=>r=3

We use this value of r:

((4),(3))(y^(4-3)(4x^3)^3)

simplifying:

((4),(3))(64yx^9)

((4),(3))=(4!)/(3!(4-3)!)=(4xx3xx2xx1)/(3xx2xx1xx1)=(4xxcancel(3)xxcancel(2)xxcancel(1))/(cancel(3)xxcancel(2)xxcancel(1)xx1)

=4/1=4

:.

The 4th term is therefore:

4*(64yx^9)

color(blue)(256yx^9)