====== Question 3 and 4, Exercise 4.8 ======
Solutions of Question 3 and 4 of Exercise 4.8 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

=====Question 3=====
Using the method of difference, find the sum of the series: $1+4+13+40+121+ \ldots$ to $n$ term.

** Solution. **
Let
$$ S_{n}=1+4+13+40+121+\ldots +T_{n} $$
Also
$$ S_{n}=1+4+13+40+\ldots +T_{n-1}+T_{n}. $$
Subtracting the second expression from the first expression, we have
\begin{align*}
	S_{n}-S_{n}& =1+4+13+40+121+\ldots +T_{n}  \\
	& -\left(1+4+13+40+\ldots +T_{n-1}+T_{n}\right)
\end{align*}
\begin{align*}
	\implies  0=&1+(4-1)+(13-4)+(40-13)+(121-40) \\
	& +\ldots+(T_{n}-T_{n-1})-T_{n}. \\
	\implies  0=&1+(3+9+27+81+\ldots \text { up to } (n-1) \text { terms })-T_{n}
\end{align*}
Then
\begin{align*}
	T_{n} & =1+(3+9+27+81+\ldots  \text { up to }(n-1) \text { terms }) \\
	& =1+\frac{3(3^{n-1}-1)}{3-1} \quad\left(\because S_{n}=\frac{a(r^n-1)}{r-1}\right) \\
	& =1+\frac{3^{n}-3}{2} \\
	& =\frac{2+3^{n}-3}{2} \\ \\
	& =\frac{3^{n}-1}{2}.
\end{align*}
Thus, the kth term of the series:
$$ T_{k}=\frac{3^{k}-1}{2}. $$
Now taking the sum, we get
\begin{align*}
	S_{n} & =\sum_{k=1}^{n} T_{k}=\sum_{k=1}^{n} \frac{3^{k}-1}{2} \\
	& =\frac{1}{2}\left(\sum_{k=1}^{n} 3^{k} - \sum_{k=1}^{n} 1\right) \\
	& =\frac{1}{2}\left((3+3^2+3^3+\ldots +3^{n})-n\right) \\
	& =\frac{1}{2}\left(\frac{3(3^n-1)}{3-1}-n\right) \\
	& =\frac{1}{2}\left(\frac{3^{n+1}-3}{2}-n\right) \\
	& =\frac{1}{4}(3^{n+1}-3-2n).
\end{align*}
Hence, the sum of the given series is \( \frac{1}{4}(3^{n+1}-3-2n) \).





=====Question 4=====
Using the method of difference, find the sum of the series: $1+2+4+7+11+16+\ldots$ to $n$ term.

** Solution. **

Let
$$ S_{n}=1+2+4+7+\ldots +T_{n} $$
Also
$$ S_{n}=1+2+4+7+\ldots +T_{n-1}+T_{n}. $$
Subtracting the second expression from the first expression, we have
\begin{align*}
S_{n}-S_{n}& =1+2+4+7+\ldots +T_{n}  \\
& -\left(1+2+4+7+\ldots +T_{n-1}+T_{n}\right)
\end{align*}
\begin{align*}
\implies  0=&1+(2-1)+(4-2)+(7-4)+\ldots+(T_{n}-T_{n-1})-T_{n}. \\
\implies  0=&1+(1+2+3+ \ldots \text{ up to } (n-1) \text{ terms })-T_{n}.
\end{align*}
Then
\begin{align*}
T_{n} & =1+(1+2+3+\ldots \text{ up to }(n-1) \text{ terms }) \\
& =1+\frac{n-1}{2}[2(1)+(n-1-1)(1)] \quad \left(\because S_n=\frac{n}{2}[2a + (n-1)d]\right) \\
& =1+\frac{(n-1)n}{2}. \\
& =\frac{2+n^2-n}{2}. \\
& =\frac{1}{2}(n^2-n+2).
\end{align*}
Thus, the \(k\)-th term of the series is:
$$ T_{k}=\frac{1}{2}(k^2-k+2). $$
Now, taking the sum, we get
\begin{align*}
S_{n} & =\sum_{k=1}^{n} T_{k}=\sum_{k=1}^{n} \left(\frac{1}{2}(k^2-k+2)\right) \\
& = \frac{1}{2} \sum_{k=1}^{n} k^2 - \frac{1}{2} \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1\\
& =\frac{1}{2} \left(\frac{n(n+1)(2n+1)}{6}\right) - \frac{1}{2} \left(\frac{n(n+1)}{2}\right)+n \\
& =\frac{n}{12}\left((n+1)(2n+1)-3(n+1)+12 \right) \\
& =\frac{n}{12}\left(2n^2+2n+n+1-3n-3+12 \right) \\
& =\frac{n}{12}\left(2n^2+10 \right) \\
& =\frac{n}{6}\left(n^2+5 \right) \\
\end{align*}
Thus
$$ S_{n} = \frac{n}{6}\left(n^2+5 \right). $$








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