Proof to the formula
1+x+x2+x3+⋯+xn=xn+1−1x−1.
Answer
Let S=1+x+x2+...+xn. Then, xS=x+x2+...+xn+1=1+x+x2+...+xn+(xn+1−1)=S+xn+1−1. So, xS−S=xn+1−1. So, S=xn+1−1x−1. (The exponent of the x in the numerator of the RHS should be n+1 not n).
Proof to the formula
1+x+x2+x3+⋯+xn=xn+1−1x−1.
Answer
Let S=1+x+x2+...+xn. Then, xS=x+x2+...+xn+1=1+x+x2+...+xn+(xn+1−1)=S+xn+1−1. So, xS−S=xn+1−1. So, S=xn+1−1x−1. (The exponent of the x in the numerator of the RHS should be n+1 not n).
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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