# Question 7 & 8 Exercise 4.3

Solutions of Question 7 & 8 of Exercise 4.3 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.

Find: $1+3-5+7+9-11+13+15-$ $17+\ldots$ upto $3 n$ terms.

We can reform the given series into three arithmetic series \begin{align}&(1+7+13+\ldots)+(3+9+15+\ldots)- \\ & (5+11+17+\ldots) \ldots \ldots \ldots . . .(1)\end{align} Now the series each one in parenthesis is arithmetic,
we can find the sum of $\mathrm{n}$ terms of each, and
then adding the $n$ terms sum of each one will give us sum of the $3 n$ terms of the given series.
For $$1+7+13+\ldots$$ here $$a_1=1, d=7-1=6$$ then sum of the $n$ terms
\begin{align}S_n&=\dfrac{n}{2}[2 a_1+(n-1) d] \text{is:}\\ S_n&=\dfrac{n}{2}[2.1+(n-1) 6]\\ \Rightarrow S_n&=\dfrac{n(6 n-4)}{2}\\ &=n(3 n-2)\end{align}
Now for $3+9+15+\ldots$, with $a_1=3, d=9-3=6$ then sum of the $\mathrm{n}$ terms, let denote
\begin{align}S_n^{\prime}&=\dfrac{n}{2}[2 a_1+(n-1) d] \text{then}\\ \cdot S_n^{\prime}&=\dfrac{n}{2}[2.3+(n-1) 6]\\ \Rightarrow S_n^{\prime}&=\dfrac{6 n^2}{2}=3 n^2\end{align}
and for $5+11+17+\ldots$ with $a_1=5$ and $d=11-5=6$ then sum of the $n$ terms,
let say denote by $S_n^{\prime \prime}$, then
$$S_n^{\prime \prime}=\dfrac{n}{2}[2 a_1+(n-1) d]$$ in this case
\begin{align}S_n^{\prime \prime}&=\dfrac{n}{2}[2.5+(n-1) 6]\\ \Rightarrow S_n^{\prime \prime}&=\dfrac{n(6 n+4)}{2}=n(3 n+2) \text {. }\end{align} Putting (i),(ii) and (iii) in (1), we get
\begin{align}S_{3 n}&=S_n+S_n^{\prime}-(S_n^{\prime \prime}) \\ \Rightarrow S_{3 n}&=n(3 n-2)+3 n^2-n(3 n+2) \\ \Rightarrow S_{3 n}&=3 n^2-2 n+3 n^2-3 n^2-2 n \\ \Rightarrow S_{3 n}&=3 n^2-4 n \\ \Rightarrow S_{3 n}&=n(3 n-4)\end{align} is the required sum of $3 n$ terms.

Show that the sum of the first $n$ odd positive integers is $n^2$.

The first odd positive in-tegers are $1,3,5, 7, \ldots$ is an arithmetic sequence
with first term $a_1=1$, and the common difference $d=2$.
We know that: \begin{align}S_n&=\dfrac{n}{2}[2 a_1+(n-1) d] \\ \therefore S_n&=\dfrac{n}{2}[2 \cdot 1+(n-1) \cdot 2] \\ \Rightarrow S_n&=\dfrac{n}{2}(2 n-2+2) \\ \Rightarrow S_n&=\dfrac{2 n^2}{2}=n^2.\end{align}