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Question: What is the next term to this series, \(2,3,7,16,32, \,and \,57....\) ?...

What is the next term to this series, 2,3,7,16,32,and57....2,3,7,16,32, \,and \,57.... ?

Explanation

Solution

For finding the next term of the series we have observed the pattern , here we have a pattern that is on adding consecutive square numbers to the preceding term , we get the next term very easily .

Complete step by step solution:
we rely on adding consecutive square numbers, the formula for the nthnth term can be obtained using the method of differences between terms ,
2 3 7 16 32 572{\text{ }}3{\text{ }}7{\text{ }}16{\text{ }}32{\text{ }}57 gives
2+(1)2=3 3+(2)2=7 7+(3)2=16 16+(4)2=32 32+(5)2=57  2 + {(1)^2} = 3 \\\ 3 + {(2)^2} = 7 \\\ 7 + {(3)^2} = 16 \\\ 16 + {(4)^2} = 32 \\\ 32 + {(5)^2} = 57 \\\
1 4 9 16 251{\text{ }}4{\text{ }}9{\text{ }}16{\text{ }}25 - the pattern of square numbers
This result in 3 5 7 93{\text{ }}5{\text{ }}7{\text{ }}9 for the differences between these square numbers, And finally, 2 2 22{\text{ }}2{\text{ }}2 .
When the third set of differences is constant, the highest term is n3{n^3} and therefore the number of n3{n^3} is that this number divided by 66 .
So the formula starts 2n36\dfrac{{2{n^3}}}{6} or n33\dfrac{{{n^3}}}{3} ,
If you then subtract n33\dfrac{{{n^3}}}{3} from the first sequence you'll be able to repeat this process to seek out the n2{n^2} component so on..
The full formula for the nthnth term is
(2n33n2+n+12)6\dfrac{{\left( {2{n^3} - 3{n^2} + n + 12} \right)}}{6}
So the 7th7th term is
(2n33n2+n+12)6\Rightarrow \dfrac{{\left( {2{n^3} - 3{n^2} + n + 12} \right)}}{6}
=(2(7)33(7)2+7+12)6 =2×3433×49+7+126 =686147+7+126 =5586 =93  = \dfrac{{\left( {2{{(7)}^3} - 3{{(7)}^2} + 7 + 12} \right)}}{6} \\\ = \dfrac{{2 \times 343 - 3 \times 49 + 7 + 12}}{6} \\\ = \dfrac{{686 - 147 + 7 + 12}}{6} \\\ = \dfrac{{558}}{6} \\\ = 93 \\\
This may look like time taking compared to noticing you’re adding a square number anytime, but if you had to seek out the 100th100th, the generic formula eventually becomes quicker.
For the record, the 100th100th term would be ,
(2n33n2+n+12)6\Rightarrow \dfrac{{\left( {2{n^3} - 3{n^2} + n + 12} \right)}}{6}
=(2(100)33(100)2+100+12)6 =2×10000003×10000+100+126 =200000030000+100+126 =19971126 =332852  = \dfrac{{\left( {2{{(100)}^3} - 3{{(100)}^2} + 100 + 12} \right)}}{6} \\\ = \dfrac{{2 \times 1000000 - 3 \times 10000 + 100 + 12}}{6} \\\ = \dfrac{{2000000 - 30000 + 100 + 12}}{6} \\\ = \dfrac{{1997112}}{6} \\\ = 332852 \\\
Note: Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily. For solving this type of question, you must understand the pattern and then try to form a general formula so that we can get any term without depending on previous output.