Give the term involving y^8 in the expansion of (2x-y^2)^10?

Question

^ means to the exponent of..

Answer 1

To get y^8 you need to have (y^2)^4

So then you do the binomial theorem:

10 C 6*(2x)^6*(-y^2)^4

[10*9*8*7/ (4*3*2*1)]*64* x^6*y^8
210*64*x^6*y^8
13440x^6y^8

Answer 2

In the expansion of (a + b)ⁿ

T_(r + 1) = nCr a^(n - r)b^r

So in the expanxion of (2x - y²)^10

T_(r + 1) = 10Cr (2x)^(10 - r)(-y²)^r

= (-1)^r 10Cr 2^(10 - r) x^(10 - r) y^(2r)

So 2r = 8, ie r = 4

So T_5 = (-1)^4 10C4 * 2^6 x^(6)y^(8)

= 1 * 210 * 64 x^(6)y^(8)
= 13440x^(6)y^(8)

Answer 3

That would be the fifth term.
The first term has y^0; the second has y^2; the third has y^4; the fourth has y^6; the fifth has y^8.
In general, the kth term (if I remember correctly) looks like this:
C(10, k-1) ((2x)^(10-k-1))((-y)^(2(k-1))
So that would be
C(10,4)((2x)^6)(-y)^(8)=
210(64x^6)(y^8) = 13440x^6y^8

Answer 4

10=X-7