urgent pattern i can not solve

• Mar 12th 2009, 12:53 AM
kl050196
urgent pattern i can not solve
What number comes next in this series????? 3, 10, 21, 44, 87
a)112
b)125
c)130
d)163
e)158
• Mar 12th 2009, 02:20 AM
Quote:

Originally Posted by kl050196
What number comes next in this series????? 3, 10, 21, 44, 87
a)112
b)125
c)130
d)163
e)158

are you sure about the correctness of this sequence
• Mar 12th 2009, 02:28 AM
kl050196
Quote:

are you sure about the correctness of this sequence

yea
• Mar 12th 2009, 03:05 AM
mr fantastic
Quote:

Originally Posted by kl050196
What number comes next in this series????? 3, 10, 21, 44, 87
a)112
b)125
c)130
d)163
e)158

Clearly option e) 158.

The sequence of numbers is obviously generated by the function $\displaystyle y = \frac{4}{3} x^3 -6x^2 + \frac{47}{3} x - 8$ operating on the integers 1, 2, 3, 4, 5, 6 ....

f(6) = 158.
• Mar 12th 2009, 05:57 AM
Soroban
Hello, kl050196!

Mr. Fantastic did his usual excellent job.
I too enjoy cranking out the generating functions for these problems.

However, since it asked for the next term (only),
. . we can use a somewhat intuitive approach.

Quote:

What number comes next in this series: .$\displaystyle 3, 10, 21, 44, 87$

. . $\displaystyle (a)\;112\qquad(b)\;125\qquad(c)\;130\qquad(d)\;163 \qquad(e)\;158$

Take the difference of consecutive terms.
Then take the differences of the differences, and so on.

$\displaystyle \begin{array}{cccccccccccc} \text{Sequence} & 3 &&10&&21&&44&&87 \\ \text{1st diff.} & & 7 && 11&&23&&43 \\ \text{2nd diff.} & & & 4 &&12&&20 \\ \text{3rd diff.} & & & & 8 && 8 \end{array}$

It seems that the 3rd differences are constant.

If this is true, we can extend the diagram to the right . . .

We assume that the next 3rd difference is also 8:

. . $\displaystyle \begin{array}{cccccccccccc} 3 &&10&&21&&44&&87 \\ & 7 && 11&&23&&43 \\ & & 4 &&12&&20 \\ & & & 8 && 8 && {\color{red}8}\end{array}$

Then the next 2nd difference must be 28:

. . $\displaystyle \begin{array}{cccccccccccc} 3 &&10&&21&&44&&87 \\ & 7 && 11&&23&&43 \\ & & 4 &&12&&20 && {\color{red}28}\\ & & & 8 && 8 && 8\end{array}$

Then the next 1st difference must be 71:

. . $\displaystyle \begin{array} {cccccccccccc} 3 &&10&&21&&44&&87 \\ & 7 && 11&&23&&43 && {\color{red}71}\\ & & 4 &&12&&20 &&28\\ & & & 8 && 8 && 8\end{array}$

Finally, the next term of the sequence must be 158:

. . $\displaystyle \begin{array} {cccccccccccc} 3 &&10&&21&&44&&87&&{\color{red}158} \\ & 7 && 11&&23&&43 && 71\\ & & 4 &&12&&20 &&28\\ & & & 8 && 8 && 8\end{array}\quad\hdots$ ta-DAA!