I am learning some series as a naive and I came across this series where I want to find the nth term.
2, 6, 14, 30, 62, ... and so on.
I am facing difficulties as the difference between elements is formed as
4, 8, 16, 32,....
Is it possible to find nth term for this series?
$\endgroup$ 52 Answers
$\begingroup$$$U_2-U_1=2^2$$
$$U_3-U_2=2^3$$
$$U_4-U_3=2^4$$
So by proceeding in this manner we get that, $$U_n-U_{n-1}=2^n$$
Now by adding these equations we get
$U_n-U_1= 2^2+2^3+2^4+.....+2^{n-1}+2^n \Rightarrow U_n=U_1+2^2+2^3+2^4+.....+2^{n-1}+2^n$
$U_n=2^1+2^2+2^3+2^4+.....+2^{n-1}+2^n$
Therefore $$U_n=\sum_{k=1}^{n}2^k$$
$$U_n=\frac{2^{n+1}-2}{2-1}$$
$$U_n=2^{n+1}-2$$
$\endgroup$ 1 $\begingroup$The first one has a logic as n+1= 2 n+2 example 6=2.2+2 14= 6.2+2 an the next one is geometric series having a common ratio 2 the nth term of this series is given by a.r^(n-1). Here a is first term so a=4 and r is the common ratio which is 2. thus nth term of 1 st sequence is n+1=2n+2 and 2nd is a.r^(n-1)
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