Question 1 Exercise 4.5

Solutions of Question 1 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Compute the sum $3+6+12+\ldots+3.2^9$

In the given geometric series: $a_1=3, \quad r=\dfrac{6}{3}=2$ and $a_n=3.2^9$.
We first find $n$ and then the sum of series.
We know that $$a_n=a_1 r^{n-1}$$,
\begin{align}3.2^9&=3(2)^{n-1} \text { or }(2)^{n-1}=\dfrac{3.2^9}{3} \\ \Rightarrow(2)^{n-1}&=2^9 \\ \Rightarrow n-1&=9 \text { or } n=10 \\ \text {. Now }\quad S_n&=\dfrac{a_1(r^n-1)}{r-1},\end{align} becomes in the given case
\begin{align}S_{10}&=\dfrac{3[2^{10}-1]}{2-1} \\ \Rightarrow \quad S_{10}&=3(2^{10}-1)\end{align}
is the required sum.

Compute the sum $8+4+2+\ldots+\dfrac{1}{16}$

In the give geometric series $$a_1=8, \quad r=\dfrac{4}{8}=\dfrac{1}{2} \quad$$
and $$a_n=\dfrac{1}{16}$$.
We first find $n$ and then the sum of series. We know that
\begin{align}a_n&=a_1 r^{n-1} \text {, }\\ \therefore \dfrac{1}{16}&=8(\dfrac{1}{2})^{n-1} or\\ (\dfrac{1}{2})^{n-1}&=\dfrac{1}{8 \times 16}\\ \Rightarrow(\dfrac{1}{2})^{n-1}&=(\dfrac{1}{2})^7\\ \Rightarrow n-1&=7\quad or \quad n=8.\\ S_n&=\dfrac{a_1(r^n-1)}{r-1} \text {, }\end{align} becomes in the given case
\begin{align}S_8&=\dfrac{8[1-(\dfrac{1}{2})^8]}{1-\dfrac{1}{2}} \\ \Rightarrow S_8&=2.8[1-\dfrac{1}{2^8}]\\ \Rightarrow S_8&=16[\dfrac{2^8-1}{2^8}] \\ \Rightarrow S_3&=\dfrac{2^8-1}{16}=\dfrac{255}{16}\end{align} is the required sum.

Compute the sum $2^4+2^5+2^6+\ldots+2^{10}$

In the give geometric series.
$$a_1=2^4, \quad r=\dfrac{2^5}{2^4}=2$$ and $$a_n=2^{10}$$.
We first find $n$ and then the sum of series. We know that
\begin{align}a_n&=a_1 r^{n-1}, \\ \therefore 2^{10}&=2^4(2)^{n-1} \text { or } 2^{n-1}=\dfrac{2^{10}}{2^4} \\ \Rightarrow 2^{n-1}&=2^6 \\ \Rightarrow n-1&=6 \text { or } n=7.\end{align} Now $$S_7=\dfrac{a_1(r^n-1)}{r-1}$$,
becomes in the given case
\begin{align}\Rightarrow S_7&=\dfrac{2^4[2^7-1]}{2-1} \\ \Rightarrow S_7&=16(512-1)=2032.\end{align}

Compute the sum $\dfrac{8}{5},-1, \dfrac{5}{8}, \ldots$,

Here $$a_1=\dfrac{8}{5}$$ \begin{align}r&=\dfrac{-1}{\dfrac{8}{5}}=-\dfrac{5}{8} \\ \therefore S_{\infty}&=\dfrac{a_1}{1-r}=\dfrac{\dfrac{8}{5}}{1-(-\dfrac{5}{8})} \\ \Rightarrow S_{\infty}&=\dfrac{8}{5(1+\dfrac{5}{8})} \\ \Rightarrow S_{\infty}&=\dfrac{8}{5} \times \dfrac{8}{13}=\dfrac{64}{65} .\end{align}

Compute the sum $2, \dfrac{2}{\sqrt{2}}, 1, \dfrac{1}{\sqrt{2}}, \dfrac{1}{2}, \ldots$,

Here $$a_1=2$$ \begin{align}r&=\dfrac{\dfrac{2}{\sqrt{2}}}{2}=\dfrac{1}{\sqrt{2}}.\\ S_{\infty}&=\dfrac{a_1}{1-r}=\dfrac{2}{1-\dfrac{1}{\sqrt{2}}}\\ \Rightarrow S_{\infty}&=\dfrac{2 \sqrt{2}}{\sqrt{2}-1}\end{align}

Compute the sum $-\dfrac{1}{3}, \dfrac{1}{2},-\dfrac{3}{4}, \ldots$, to $7$ terms.

Here $$a_1=-\dfrac{1}{3}$$ and
\begin{align}r&=\dfrac{\dfrac{1}{2}}{-\dfrac{1}{3}}=\dfrac{3}{2} . \\ S_7&=\dfrac{a_1(1-r^7)}{1-r} \text { becomes } \\ S_7&=\dfrac{-\dfrac{1}{3}(1+\dfrac{1}{3^7})}{\dfrac{4}{3}} \\ S_7&=-\dfrac{1}{3} \times \dfrac{2188}{2187} \times \dfrac{3}{4} \\ S_7&=\dfrac{547}{2187}\end{align}

Question 2 >