Question

Find the $n^{th}$ term and sum to n terms of the series:
$\begin{aligned} & 1+5+13+29+61+........ \\\ & a){{T}_{n}}={{2}^{n+1}}-3 \\\ & \,\,\,\,{{S}_{n}}=({{2}^{2}}+{{2}^{3}}+............+{{2}^{n+1}})-3n \\\ & b){{T}_{n}}={{2}^{n+1}}+3 \\\ & \,\,\,\,{{S}_{n}}=({{2}^{2}}+{{2}^{3}}+............+{{2}^{n+1}})+3n \\\ & c){{T}_{n}}={{2}^{n+1}}-3 \\\ & \,\,\,\,{{S}_{n}}=({{2}^{2}}+{{2}^{3}}+............-{{2}^{n+1}})-3n \\\ & d){{T}_{n}}={{2}^{n-1}}-3 \\\ & \,\,\,\,{{S}_{n}}=({{2}^{2}}+{{2}^{3}}+............+{{2}^{n+1}})+3n \\\ \end{aligned}$

Answer 1

Write the given terms as:
$\begin{aligned} & 1=({{2}^{2}}-3);\,5=({{2}^{3}}-3);\,13=({{2}^{4}}-3);\,29=({{2}^{5}}-3) \\\ & \\\ \end{aligned}$
and so on. Observe the obtained pattern and write the $n^{th}$ term (Tn) of the series. Now to write the expression of the sum of these ‘n’ terms of the series, group the terms containing exponents of $2$ together and the constant term $3$ together. Find the sum of these ‘n’ $3$’s and get the answer.

Complete step by step answer:
Here we have been provided with the series: $1+5+13+29+61+..........n\,\text{terms}$. We have been asked to determine the expression for $n^{th}$ term and sum of these n terms of the series.
Let us denote given term as ${{T}_{1}},{{T}_{2}},{{T}_{3}},.............$ and so on where ${{T}_{1}}$ represents first term ${{T}_{2}}$ represents second term, ${{T}_{3}}$ represents third term and so on. So, we can write the given terms as:

& {{T}_{1}}=1={{2}^{2}}-3={{2}^{1+1}}-3 \\\ & {{T}_{2}}=5={{2}^{3}}-3={{2}^{2+1}}-3 \\\ & {{T}_{3}}=13={{2}^{4}}-3={{2}^{3+1}}-3 \\\ & {{T}_{4}}=29={{2}^{5}}-3={{2}^{4+1}}-3 \\\ & {{T}_{5}}=61={{2}^{6}}-3={{2}^{5+1}}-3 \\\ \end{aligned}$$ So, on observing the above pattern we can write the expression for $n^{th}$ term by Tn as: $\Rightarrow {{T}_{n}}={{2}^{n+1}}-3........(i)$ Now, let us find the expression for the sum of these ‘n’ terms. So, we have, $\Rightarrow {{S}_{n}}({{2}^{2}}-3)+({{2}^{3}}-3)+.............+({{2}^{n+1}}-3)$ So, grouping the terms containing exponents of $2$ together and all the $3$’s together, we have, $\Rightarrow {{S}_{n}}({{2}^{2}}+{{2}^{3}}+.............+{{2}^{n+1}})-(3+3+3+..........n\,\text{times})$ Clearly, we can see that the constant term $3$’s is added n times so its sum will be the product of $3$ and n i.e. $3$n so, we get, $\Rightarrow {{S}_{n}}({{2}^{2}}+{{2}^{3}}+.............+{{2}^{n+1}})-3n..........(ii)\,$ From the obtained expression of Tn and Sn in equations (i) and (ii) respectively, **we have the conclusion that option (a) is the correct answer.** **Note:** One may note that in the expression of Sn we do not have to find the simplified sum of the expression $({{2}^{2}}+{{2}^{3}}+.............+{{2}^{n+1}})$because in the options we aren’t provided the simplified form. Although if you want to simplify the expression $({{2}^{2}}+{{2}^{3}}+.............+{{2}^{n+1}})$ then apply the formula given as $S=\dfrac{a({{r}^{n}}-1)}{r-1}$, where ‘a’ is the first term ‘r’ is the common ratio and ‘s’ denotes the sum of ‘n’ terms of G.P.