Sum of the first n terms of a G.P. (Sn)

Consider the G.P. $t_1,t_2,t_3,\ldots,t_n,\ldots$; the sum of its first $n$ terms is written $S_n=\sum_{r=1}^{n}t_r=t_1+t_2+\cdots+t_n$. Here $\sum$ is the notation for summation and $r$ is the running (dummy) variable.

Theorem: for a G.P. $a,ar,ar^2,\ldots,ar^{n-1}$ with $r\ne1$, $S_n=\dfrac{a(r^n-1)}{r-1}$.

Proof: $S_n=a+ar+ar^2+\cdots+ar^{n-1}=a(1+r+r^2+\cdots+r^{n-1})\ \ldots(1)$. Multiplying both sides by $r$: $rS_n=a(r+r^2+r^3+\cdots+r^n)\ \ldots(2)$. Subtracting (2) from (1): $S_n-rS_n=a(1-r^n) \Rightarrow S_n(1-r)=a(1-r^n) \Rightarrow S_n=\dfrac{a(1-r^n)}{1-r}$, $r\ne1$.

Let's Note: (1) if $0<r<1$ it is convenient to write $S_n=\dfrac{a(1-r^n)}{1-r}$; (2) if $r>1$ it is convenient to write $S_n=\dfrac{a(r^n-1)}{r-1}$; (3) if $r=1$, the G.P. is $a,a,a,\ldots,a$ ($n$ times), so $S_n=na$; (4) $S_n-S_{n-1}=t_n$ (removing the sum of the first $n-1$ terms from the sum of the first $n$ terms leaves exactly the $n$th term -- useful for recovering $t_n$ when only $S_n$ is given).

Worked Example 4: for a G.P. $a=6, r=2$, find $S_{10}$. $S_{10}=\dfrac{6(2^{10}-1)}{2-1}=6(1023)=6138$.

Worked Example 5: how many terms of $2,2^2,2^3,2^4,\ldots$ are needed to give the sum 2046? $2046=\dfrac{2(2^n-1)}{2-1}=2(2^n-1) \Rightarrow 1023=2^n-1 \Rightarrow 2^n=1024=2^{10} \Rightarrow n=10$.

Worked Example 6: for a G.P. $r=2$, $S_{10}=1023$, find $a$. $1023=a\cdot\dfrac{2^{10}-1}{2-1}=1023a \Rightarrow a=1$.

Worked Example 7: for a G.P. $a=3, r=2, S_n=765$, find $n$. $765=3(2^n-1) \Rightarrow 255=2^n-1 \Rightarrow 2^n=256=2^8 \Rightarrow n=8$.

Worked Example 8: for a G.P. $S_3=16, S_6=144$, find $a$ and $r$. $\dfrac{S_6}{S_3}=\dfrac{r^6-1}{r^3-1}=\dfrac{(r^3-1)(r^3+1)}{r^3-1}=r^3+1=\dfrac{144}{16}=9 \Rightarrow r^3=8 \Rightarrow r=2$. From $S_3=16$: $a\dfrac{2^3-1}{2-1}=7a=16 \Rightarrow a=\dfrac{16}{7}$.

Worked Example 9: find $5+55+555+5555+\cdots$ upto $n$ terms. $S_n=5(1+11+111+\cdots)=\dfrac59\bigl[(10-1)+(100-1)+(1000-1)+\cdots \text{ to } n \text{ brackets}\bigr]=\dfrac59\Bigl[(10+100+\cdots+10^n)-n\Bigr]=\dfrac59\left[\dfrac{10(10^n-1)}{9}-n\right]$.

Worked Example 10: find the sum to $n$ terms of $0.3+0.33+0.333+\cdots$. $S_n=3[0.1+0.11+0.111+\cdots]=\dfrac39\bigl[(1-0.1)+(1-0.01)+(1-0.001)+\cdots\bigr]=\dfrac39\bigl[n-(0.1+0.01+0.001+\cdots\text{ to }n\text{ terms})\bigr]=\dfrac13\left[n-\dfrac{0.1(1-0.1^n)}{1-0.1}\right]=\dfrac n3-\dfrac{1-0.1^n}{27}$.

Worked Example 11: find the $n$th term of $0.4,0.44,0.444,\ldots$. $t_1=0.4,\ t_2=0.4+0.04,\ t_3=0.4+0.04+0.004,\ldots$, so $t_n$ is itself the sum of the first $n$ terms of a G.P. with $a=0.4,r=0.1$: $t_n=\dfrac{0.4(1-0.1^n)}{1-0.1}=\dfrac49\bigl[1-(0.1)^n\bigr]$.

Worked Example 12: for a sequence $S_n=5(4^n-1)$, find $t_n$, verify it is a G.P., and find $r$. $t_n=S_n-S_{n-1}=5(4^n-1)-5(4^{n-1}-1)=5(4^n-4^{n-1})=5\cdot4^{n-1}(4-1)=15(4^{n-1})$. Then $\dfrac{t_{n+1}}{t_n}=\dfrac{15\cdot4^n}{15\cdot4^{n-1}}=4$, a constant, so the sequence is a G.P. with $t_n=15(4^{n-1})$ and $r=4$. …