What's the answer for the sum with $$ \sum_{r=0}^n \left(\frac 34 \right)^r $$
I got $4\left(1-\left(\frac 34 \right)^n\right)$ but apparently it's wrong. My work is provided in the picture uploaded.enter image description here Any help is appreciated. Thanks!
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$$\sum_{r=0}^n \left(\frac 34 \right)^r$$ is the sum of a geometric sequence, namely $$ 1+(\frac {3}{4}) +(\frac {3}{4})^2 +...+ (\frac {3}{4})^n$$
We know that $$1 + r +r^2 + ...+ r^n = \frac{r^{n+1}-1}{r-1}$$
Thus $$\sum_{r=0}^n \left(\frac 34 \right)^r = \frac{(3/4)^{n+1}-1}{(3/4)-1}= 4\left(1-\left(\frac {3}{4} \right)^{n+1}\right) $$
The formula for $1+x+\cdots + x^n$ is $\frac{1-x^{n+1}}{1-x}$. So replace $n$ by $n+1$ in your answer and you are good to go!