====== Question 5 Exercise 4.1 ======
Solutions of Question 5 of Exercise 4.1 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.

=====Question 5(i)=====
Write each of the following series in expanded form, $\sum_{j=1}^6(2 j-3)$
====Solution====
\begin{align}\sum_{j=1}^6(2 j-3)&=(2.1-3)+(2.2-3)+(2.3-3)+(2.4-3)\\&+(2.5-3)+(2.6-3) \\
\implies \sum_{j=1}^6(2 j-3)&=-1+1+3+5+7+9 .\end{align}

=====Question 5(ii)=====
Write each of the following series in expanded form, $\sum_{k=1}^5(-1)^k 2^{k-1}$
====Solution====
\begin{align}\sum_{k=1}^5(-1)^k 2^{k-1}& =(-1)^1 2^{1-1}+(-1)^2 2^{2-1}\\
&+(-1)^3 2^{3-1}+(-1)^4 2^4 1 +(-1)^5 2^{5-1} \\
\implies \sum_{k=1}^5(-1)^k 2^k& =-1+2-4+8-16.\end{align}

=====Question 5(iii)=====
Write each of the following series in expanded form, $\sum_{j=1}^{\infty} \dfrac{1}{2^j}$
====Solution====
\begin{align}\sum_{j=1}^{\infty} \dfrac{1}{2^j}&=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\ldots\\
&=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\ldots \end{align}
or we can simply write
\begin{align}\sum_{j=1}^{\infty} \dfrac{1}{2^j}=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\ldots \end{align}

=====Question 5(iv)=====
Write each of the following series in expanded form, $\sum_{k=0}^{\infty}\left(\dfrac{3}{2}\right)^k$
====Solution====
\begin{align}\sum_{k=0}^{\infty}\left(\dfrac{3}{2}\right)^k&=\left(\dfrac{3}{2}\right)^0+\left(\dfrac{3}{2}\right)^1+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\ldots \\
&=1+\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{27}{8}+\ldots\end{align}
or we can simply write
\begin{align}\sum_{k=0}^{\infty}\left(\dfrac{3}{2}\right)^k&=1+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\ldots \end{align}




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