Question
Sum of $$n$$ terms of series $$12 + 16 + 24 + 40 + . . . . .$$ will be
A.
$$2\left( {{2^n} - 1} \right) + 8n$$
B.
$$2\left( {{2^n} - 1} \right) + 6n$$
C.
$$3\left( {{2^n} - 1} \right) + 8n$$
D.
$$4\left( {{2^n} - 1} \right) + 8n$$
Answer :
$$4\left( {{2^n} - 1} \right) + 8n$$
Solution :
Let $$n^{th}$$ term of series is $$T_n$$ then
$$\eqalign{
& {S_n} = 12 + 16 + 24 + 40 + .... + {T_n} \cr
& {\text{Again }}\,{S_n} = 12 + 16 + 24 + .... + {T_n} \cr} $$
On subtraction
$$\eqalign{
& 0 = \left( {12 + 4 + 8 + 16 + .... + \,{\text{upto }}n{\text{ terms}}} \right) - {T_n} \cr
& {\text{or }}{T_n} = 12 + \left[ {4 + 8 + 16 + .... + \,{\text{upto}}\left( {n - 1} \right){\text{terms}}} \right] \cr
& = 12 + \frac{{4\left( {{2^{n + 1}} - 1} \right)}}{{2 - 1}} = {2^{n - 1}} + 8 \cr} $$
On putting $$n = 1, 2, 3 ....$$
$$\eqalign{
& {T_1} = {2^2} + 8,{T_2} = {2^3} + 8,{T_3} = {2^4} + 8.....\,{\text{etc}}{\text{.}} \cr
& {S_n} = {T_1} + {T_2} + {T_3} + .... + {T_n} \cr
& = \left( {{2^2} + {2^3} + {2^4} + ....\,{\text{upto }}n{\text{ terms}}} \right) + \left( {8 + 8 + 8 + ....\,{\text{upto }}n{\text{ terms}}} \right) \cr
& = \frac{{{2^2}\left( {{2^n} - 1} \right)}}{{2 - 1}} + 8n = 4\left( {{2^n} - 1} \right) + 8n. \cr} $$