Maharashtra BoardEnglish MediumSTD 11 ScienceMathsGeometric Progressions5 Marks
Question
Sum the following series to $n$ terms:
$1 + 3 + 7 + 13 + 21 + .....$
✓
Answer
We have,
$1 + 3 + 7 + 13 + 21 + .....$
The sequence of the differences between the successive terms of the this series is $2, 4, 6, 8 ......$ Clearly,
it is an A.P. with common difference $2.$
Let $T_n$ be the $n^{th}$ term and $S_n$ denote the sum of $n$ terms of the given series.
Then, $S_n = 1 + 3 + 7 + 13 + 21 + ...... + T_{n-1} + T_n .....(i)$
Also, $S_n = 1 + 3 + 7 + 13 + ...... + T_{n-1} + T_n ....(ii)$
Subtracting $(ii)$ from $(i),$ we get
$0=1+\big[2+4+6+8+\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]-\text{T}_\text{n}$
$\text{T}_\text{n}=1+\big[2+4+6+8+\ ....\ +(\text{T}_\text{n}-\text{T}_{\text{n}-1})\big]$
$\text{T}_\text{n}=1+\frac{(\text{n}-1)}{2}\big[2\times2+(\text{n}-1-1)\times2\big]$
$\text{T}_\text{n}=1+\frac{(\text{n}-1)}{2}\times2\big[2+(\text{n}-2)\big]$
$\text{T}_\text{n}=1+(\text{n}-1)(\text{n})$
$\text{T}_\text{n}=1+\text{n}^2-\text{n}$
$\text{T}_\text{n}=\text{n}^2-\text{n}+1$
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}=\sum\limits^{\text{n}}_{\text{k}=1}\big(\text{k}^2-\text{k}+1\big)$
$\text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2-\sum\limits^{\text{n}}_{\text{k}=1}\text{k}+\sum\limits^{\text{n}}_{\text{k}=1}1$
$=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}-\frac{\text{n}(\text{n}+1)}{2}+\text{n}$
$=\frac{\text{n}(\text{n}+1)(2\text{n}+1)-3\text{n}(\text{n}+1)+6\text{n}}{6}$
$=\frac{\text{n}}{6}\big[(\text{n}+1)(2\text{n}+1)-3(\text{n}+1)+6\big]$
$=\frac{\text{n}}{6}\big[2\text{n}^2+\text{n}+2\text{n}+1-3\text{n}-3+6\big]$
$=\frac{\text{n}}{6}\big[2\text{n}^2+4\big]$
$=\frac{\text{n}}{6}\times2\big[\text{n}^2+2\big]$
$=\frac{\text{n}}{3}(\text{n}^2+2)$
Hence, $\text{S}_\text{n}=\frac{\text{n}}{3}(\text{n}^2+2)$
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