Question

The sum of 40 terms of the series
 $1+2+3+4+5+8+7+16+9+........$ is
A. $398+2^{20}$
B. $398+2^{21}$
C. $398+2^{19}$
D. None of these

Answer 1

Hint: This problem deals with both arithmetic and geometric progression. In an arithmetic progression, the consecutive terms differ by a common difference $d$ , with an initial term $a$ at the beginning. Whereas in the geometric progression, the consecutive terms differ by a common ratio $r$ , also with an initial term $a$ at the beginning. The following formulas are used in the problem.
The sum of the $n$ terms in an A.P. is given by:
 $\Rightarrow S_n={\displaystyle \frac{n}{2}}\left[2a+\left(n-1\right)d\right]$
The sum of the $n$ terms in a G.P. is given by:
 $\Rightarrow {\displaystyle \frac{a\left(r^n-1\right)}{r-1}}$

Complete step-by-step answer:
Given that there are in total 40 terms in the series which is given by:
 $\Rightarrow 1+2+3+4+5+8+7+16+9+........$
Seems like this doesn’t follow any pattern so splitting the alternate terms of the above given series as:
 $\Rightarrow \left(3+5+7+9+....\right)+\left(1+2+4+8+16+....\right)$
The first split terms are strictly odd numbers and have a common difference and hence they are in A.P
Whereas the second arranged terms have a common ratio and are hence in a G.P.
Consider the first arrangement of terms in the series which are:
 $\Rightarrow 3+5+7+9+....$ up to 19 terms in total in this first part of the series.
The above series have a common difference of 2, and the initial term is 3 here.
Hence the sum of the terms up to 19 terms which are in A.P is given by:
 $\Rightarrow 3+5+7+9+....$ 19 terms
 $\Rightarrow {\displaystyle \frac{19}{2}}\left[2(3)+\left(19-1\right)\left(2\right)\right]$
 $\Rightarrow {\displaystyle \frac{19}{2}}\left[6+18\left(2\right)\right]$
 $\Rightarrow {\displaystyle \frac{19}{2}}\left[42\right]$
 $\Rightarrow 399$
Now consider the second part of the series which are:
 $\Rightarrow 1+2+4+8+16+....$ up to 21 terms in total in the second part of the series.
The above series have a common ratio of 2, and the initial term is 1 here.
Hence the sum of the terms up to 21 terms which are in G.P is given by:
 $\Rightarrow 1+2+4+8+16+....$ 21 terms
 $\Rightarrow {\displaystyle \frac{1\left(2^{21}-1\right)}{2-1}}$
 $\Rightarrow \left(2^{21}-1\right)$
Now adding the sum of the two parts of the series :
 $\Rightarrow \left(3+5+7+9+....\right)+\left(1+2+4+8+16+....\right)$
 $\Rightarrow 399+2^{21}-1$
 $\Rightarrow 398+2^{21}$

Final Answer: The sum of 40 terms of the given series is $398+2^{21}$

Note:
Please note that the formula used for finding the sum of $n$ terms in the G.P varies accordingly, that is if the common ratio of the G.P is greater than 1, then the formula applied in the problem is used.
But the sum of the $n$ terms in G.P if the common ratio is less than 1 which is $r<1$ , is given by:
 $\Rightarrow {\displaystyle \frac{a\left(1-r^n\right)}{1-r}}$
The sum of the infinite terms of a G.P when $\left|r\right|<1$ , is given by:
 $\Rightarrow {\displaystyle \frac{a}{1-r}}$