Some Special Series — Maths STD 11 Science — Question
CBSE BoardEnglish MediumSTD 11 ScienceMathsSome Special Series5 Marks
Question
Sum the following series to n terms:
$2 + 5 + 10 + 17 + 26 + .....$
✓
Answer
Let $T_n $ be the $n^{th} $ term and $S_n $ be the sum of n terms of the given series.
Thus, we have
$S_n = 2 + 5 + 10 + 17 + 26 + ..... + T_{n-1} + T_n ....(i)$
Equation (i) can be rewritten as,
$S_n = 2 + 5 + 10 + 17 + 26 + ...... + T_{n-1} + T_n ....(ii)$
On subtracting (ii) from (i), we get
$\underline{\underline{\text{S}_\text{n} = 2\ +\ 5\ +\ 10\ +\ 17 \ +\ 26\ +\ .........\ +\ \text{T}_{\text{n}-1}\ +\ \text{T}_\text{n}\\\text{S}_\text{n} =\ \ \ \ \ \ \ \ \ 2 \ +\ \ 5\ \ + \ 10\ +\ 17\ + 6 \ +\ ....\ \ +\ \text{T}_{\text{n}-1}\ +\ \text{T}_\text{n}}}\\\ 0\ =\ 2\ +\big[3\ +\ 5\ +\ 7\ +\ 9+\ .............\ +\ (\text{T}_\text{n}-\text{T}_{\text{n}-1})-\text{T}_\text{n}$
The sequence of difference of successive terms is 3, 5, 7, 9, .....
We observe that it is an AP with common difference 2 and first term 3.
Thus, we have
$2+\Big[\frac{(\text{n}-1)}{2}\big\{6+(\text{n}-2)2\big\}\Big]-\text{T}_\text{n}=0$
$\Rightarrow2+\big[\text{n}^2+1\big]=\text{T}_\text{n}$
$\Rightarrow\big[\text{n}^2+1\big]=\text{T}_\text{n}$
Now,
$\because\ \text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}$
$\therefore\ \text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\big(\text{k}^2+1\big)$
$\Rightarrow\text{ S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{k}^2+\sum\limits^{\text{n}}_{\text{k}=1}1$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\text{n}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)(2\text{n}+1)+6\text{n}}{6}$
$\Rightarrow\text{S}_\text{n}=\frac{\text{n}(2\text{n}^2+3\text{n}+7)}{6}$
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