Question

Sum of \[n\] terms of series will be \[12 + 16 + 24 + 40 + ...\] will be
\[2({2^n} - 1) + 8n\]
\[2({2^n} - 1) + 6n\]
\[3({2^n} - 1) + 8n\]
\[4({2^n} - 1) + 8n\]

Answer 1

Hint: Geometric Progression (GP) is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio. This progression is also known as a geometric sequence of numbers that follow a pattern.

Complete step-by-step answer:
General form of a GP
\[a,ar,a{r^2},a{r^3},...,a{r^n}\]
where \[a\] is the first term
\[r\] is the common ratio
\[{r^n}\] is the last term
If the common ratio is:
Negative: the result will alternate between positive and negative.
Greater than \[1\] : there will be an exponential growth towards infinity (positive).
Less than \[ - 1\] : there will be an exponential growth towards infinity (positive and negative).
Between \[1\] and \[ - 1\]: there will be an exponential decay towards zero.
Zero: the result will remain at zero
When three quantities are in GP, the middle one is called the geometric mean of the other two.
 Suppose \[a,ar,a{r^2},a{r^3},...,a{r^n}\] s the given Geometric Progression.
Then the sum of \[n\]terms of GP is given by:
\[S = a\left( {\dfrac{{{r^n} - 1}}{{r - 1}}} \right)\]
Let \[S = 12 + 16 + 24 + 40 + ...{t_n}\]
This can be rewritten as ,
\[S = (8 + {2^2}) + (8 + {2^3}) + (8 + {2^4}) + ... + (8 + {2^{n + 1}})\]
Taking \[8\] common from all the terms we get ,
\[ = 8n + {2^2} + {2^3} + {...2^{n + 1}}\]
Except for the first term, all the terms forms a Geometric Progression (GP) with \[{2^2}\] as first term and common ratio as \[2\]
Therefore we get ,
\[ = 8n + {2^2}\dfrac{{({2^n} - 1)}}{{2 - 1}}\]
Which simplifies to ,
\[ = 8n + 4({2^n} - 1)\]
Therefore option (4) is the correct answer.
So, the correct answer is “Option 4”.

Note: Geometric Progression (GP) is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio Then the sum of \[n\]terms of GP is given by:
\[S = a\left( {\dfrac{{{r^n} - 1}}{{r - 1}}} \right)\]