Express each sum using summation notation:

(a) $1^{2}+2^{2}+3^{2}+\cdots+9^{2}$

(b) $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots+\dfrac{1}{2^{n-1}}$

Solution (a) The sum $1^{2}+2^{2}+3^{2}+\cdots+9^{2}$ has $9$ terms, each of the form $k^{2}$; it starts at $k=1$ and ends at $k=9$ : $$1^{2}+2^{2}+3^{2}+\cdots+9^{2}=\sum\limits_{k=1}^{9}k^{2}$$

(b) The sum $$1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots+\dfrac{1}{2^{n-1}}$$ has $n$ terms, each of the form $\dfrac{1}{2^{k-1}}$. It starts at $k=1$ and ends at $k=n$. $$1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots+\dfrac{1}{2^{n-1}} =\sum\limits_{k=1}^{n}\dfrac{1}{2^{k-1}}$$