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Expand (1-1/2x)^5 Using Binominal theorem.​

User Natoya
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The binomial theorem states that for any positive integer n, the expansion of (a+b)^n can be written as the sum of the terms of the form C(n,r)*a^(n-r)*b^r, where C(n,r) denotes the binomial coefficient, or the number of ways to choose r objects from a set of n objects.

In this case, we want to expand (1-1/2x)^5. Using the binomial theorem, we have:

(1-1/2x)^5 = C(5,0)1^5(-1/2x)^0 + C(5,1)1^4(-1/2x)^1 + C(5,2)1^3(-1/2x)^2 + C(5,3)1^2(-1/2x)^3 + C(5,4)1^1(-1/2x)^4 + C(5,5)1^0(-1/2x)^5

Simplifying this expression, we get:

(1-1/2x)^5 = 1 - 5/2x + 25/8x^2 - 25/16x^3 + 625/256x^4 - 3125/1024x^5

Therefore, the expansion of (1-1/2x)^5 using the binomial theorem is 1 - 5/2x + 25/8x^2 - 25/16x^3 + 625/256x^4 - 3125/1024x^5.

User Sai Prashanth
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