How can we calculate? (1-x) + (1-x)^2 + (1-x)^3 + (1-x)^4 + (1-x)^5.... + (1-x)^n

1 Answer
Cesareo R.
Mar 24, 2018

((1 + (1 - x)^n) (1-x))/( x-2)(1+(1−x)n)(1−x)x−2

Explanation:

The polynomial identity

sum_(k=0)^n a^k = (a^(n+1)-1)/(a-1)n∑k=0ak=an+1−1a−1 leave us to

sum_(k=1)^n (1-x)^k =((1-x)^(n+1)-1)/((x-1)-1)-1 = ((1 + (1 - x)^n) (1-x))/( x-2)n∑k=1(1−x)k=(1−x)n+1−1(x−1)−1−1=(1+(1−x)n)(1−x)x−2

For x ne 2x≠2

If instead we have

sum_(k=0)^(n-1) (1-x)^k =((1-x)^n-1)/((x-1)-1) = ((1-x)^n-1)/(x-2) n−1∑k=0(1−x)k=(1−x)n−1(x−1)−1=(1−x)n−1x−2

for x ne 2x≠2

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