Question

Complete the series 4, 5, 12, 39, 160, …..?

Answer 1

Hint: For solving this question you should know about the series and its basic properties. As we know that a series is always written in a fixed manner and there will always be a common term on which they work or their next terms are written. These are always in the fixed formula which is determined by subtracting the terms from their next term and then it will be cleared.

Complete step by step solution:
According to our question we have to ask to complete the series which is given as 4, 5, 12, 39, 160, ……
So, as we know, a series is always written in a fixed manner. It means if we want to write any next element of that series then that will be written by a fixed calculation. It can’t be written randomly. And this fixed manner will be found for a sequence by difference finding between two successive terms and these will provide us a fixed manner in which we will write the next term. And it is always compulsory if there is a series unless it can’t be a series.
So, if we have to find the next terms of a given series then first we will find the fixed rule of the series or the fixed manner for the series from which we are going to count our next element of this series.
So, here if we see for the fixed rule, then:
\[\Rightarrow 5-4=1,\Rightarrow 12-5=7,\Rightarrow 39-12=27\]
So, here we can see that there is no common difference and by these the rule will be,
\[\left[ {{a}_{\left( n-1 \right)}}*\left( n-1 \right) \right]+\left( n-1 \right)\]
If we check our rule, then for \[{{2}^{nd}}\] element of series \[n=2\];
\[=\left[ {{a}_{\left( 2-1 \right)}}*1 \right]+1=\left( 4*1 \right)+1=5\]
For \[n=3;=\left( {{a}_{3-1}}\times 2 \right)+2=\left( 5\times 2 \right)+2=12\]
For \[n=4;=\left( {{a}_{4-1}}\times 3 \right)+3=\left( 12\times 3 \right)+3=39\]
For \[n=5;=\left( {{a}_{5-1}}\times 4 \right)+4=\left( 39\times 4 \right)+4=160\]
For \[n=6;=\left( {{a}_{6-1}}\times 5 \right)+5=\left( 160\times 5 \right)+5=805\]
For \[n=7;=\left( {{a}_{7-1}}\times 6 \right)+6=\left( 805\times 6 \right)+6=4836\]
For \[n=8;=\left( {{a}_{8-1}}\times 7 \right)+7=\left( 4836\times 7 \right)+7=33859\]
So, the series will be 4, 5, 12, 39, 160, 805, 4836, …….

Note: While solving this problem, you always have to check the rule which you have to find by the given terms and then compare the new terms with the given original terms, if it will be the same then our rule is right unless the rule will be wrong.