how do you use Pascal's Triangle in this equation: (5-2y^2)^5 ?

Answer 1

The coefficients of the binomial expansion for (a+b)^n correspond to the numbers on the nth row of "my" triangle (remembering that the top row, consisting of a single 1, is row 0 by convention). So first, write out the rows of the triangle up to the fifth:

....... 1
....... 1 1
.....1 .2 1
..1 .3 ..3 1
1 .4 .6 ..4 1
1 5 10 10 5 1

So the binomial expansion of (a+b)⁵ is simply:

a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵

Now just substitute a=5, b=-2y²:

5⁵ - 5*5⁴*2y² + 10*5³*4y⁴ - 10*25*8y⁶ + 5*5*16y⁸ - 32y¹⁰

And simplify:

3125 - 6250y² + 5000y⁴ - 2000y⁶ + 400y⁸ - 32y¹⁰

And we are done.

Answer 2

(a+b)^5 =(a+b)^2*(a+b)^2*(a+b) =(a^2+2a*b+b^2)*(a^2+2a*b+b^2)*(a+b) =(a^4+2a^3*b+a^2*b^2+2a^3*b+4a^2*b^2+2... =(a^4+4a^3b+6a^2b^2+4ab^3+b^4)*(a+b) =(a^5+a^4*b+4a^4*b+4a^3*b^2+6a^3*b^2+6... =a^5+5a^4*b+10a^3*b^2+10a^2*b^3+5a*b^4...