I have a vague idea about how to solve (a+3b)^5, but get lost in the second set of equations... I just need to see it done to understand it, I think.
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Since (x + y)^5
= C(5, 0) x^5 + C(5, 1) x^4 y + C(5, 2) x^3 y^2 + C(5, 3) x^2 y^3 + C(5, 4) xy^4 + C(5, 5) y^5
= x^5 + 5x^4 y + 10x^3 y^2 + 10x^2 y^3 + 5xy^4 + y^5, by Pascal's Triangle or otherwise:
letting x = a, y = 3b yields (a + 3b)^5
= a^5 + 5a^4 (3b) + 10a^3 (3b)^2 + 10a^2 (3b)^3 + 5a(3b)^4 + (3b)^5
= a^5 + 15a^4 b + 90a^3 b^2 + 270a^2 b^3 + 405ab + 243b^5.
I hope this helps!
= C(5, 0) x^5 + C(5, 1) x^4 y + C(5, 2) x^3 y^2 + C(5, 3) x^2 y^3 + C(5, 4) xy^4 + C(5, 5) y^5
= x^5 + 5x^4 y + 10x^3 y^2 + 10x^2 y^3 + 5xy^4 + y^5, by Pascal's Triangle or otherwise:
letting x = a, y = 3b yields (a + 3b)^5
= a^5 + 5a^4 (3b) + 10a^3 (3b)^2 + 10a^2 (3b)^3 + 5a(3b)^4 + (3b)^5
= a^5 + 15a^4 b + 90a^3 b^2 + 270a^2 b^3 + 405ab + 243b^5.
I hope this helps!