expansion of{ x+a}{ 2y^5}
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(1+x)^n = 1 + nx + [n(n-1)/2!]x^2 + [n(n-1)(n-2)/3!]x^3 + [n(n-1)(n-2)(n-3)/4!]x^4 + [n(n-1)………1)/n!]x^n
(x+2y)^5 = [x{1+(2y/x)}]^5 = x^5 [1 + (2y/x)]^5
(x+2y)^5 = x^5[1 + 5(2y/x) + (5.4)/2! (2y/x)^2 + (5.4.3)/3! (2y/x)^3 + (5.4.3.2)/4! (2y/x)^4 + (5.4.3.2.1)/5! (2y/x)^5
(x+2y)^5 = x^5[1 + 10y/x + 10(4y^2/x^2) + 10(8y^3/x^3) + 5(16y^4/x^4) + (32y^5/x^5)
(x+2y)^5 = x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80 xy^4 + 32y^5.
its prob confusing but tried my best last answer was prob wrong
Answered by
0
Step-by-step explanation:
= x(2y^5)+a (2y^5)
= 2xy^5 + 2ay^5
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