hi i need a recursive formula for these numbers 3,10,21, 36...,i need to show how to get to the next #?

Question

i need to find a way to relate each number to the previous number...like a(sub n-1) is related to (a sub n) how?

Answer 1

55, 78, 105, 136, 171, 210, 253, 300, 351, 406, 465, 528 ...

It is a matter of noticing the pattern, that the difference between successive numbers increments by 4 each time:

10 - 3 = 7
21 - 10 = 11
36 - 21 = 15

So to get the next number we add 19:

36 + 19 = 55

then we add 23:

55 + 23 = 78
78 + 27 = 105
105 + 31 = 136
136 + 35 = 171
171 + 39 = 210
210 + 43 = 253
253 + 47 = 300
300 + 51 = 351
351 + 55 = 406
406 + 59 = 465
465 + 63 = 528

Thus a (sub n) = (4n - 1) + a (sub n-1)

a (sub 2) = 7 + a (sub 1)
a (sub 3) = 11 + a (sub 2)
a (sub 4) = 15 + a (sub 3) etc

Answer 2

Relation between the successive numbers is as folllows :-
3
3+7=10
10+(7+4)=10+11=21
21+(11+4)=21+15=36
36+(15+4)=36+19=55

Answer 3

Let's see if there's a pattern in the differences...

10 - 3 = 7
21 - 10 = 11
36 - 21 = 15

Each difference is four more than the difference before it. So, two questions now: (1) what's the next number in the sequence? and (2) how can we characterize this sequence in a formula?

(1) The next difference should be 19, and 36 + 19 = 55, so 55 is the next number in the sequence.

(2) Let's call the nth entry in the sequence a_n. a_1 = 3, of course. Then we have

a_2 = a_1 + 7
a_3 = a_2 + 11
a_4 = a_3 + 15
a_5 = a_4 + 19

But since we want the differences (7, 11, 15, 19, etc.) to be expressed in terms of their step-by-step growth, each step growing by 4, let's write them this way:

a_2 = a_1 + 3 + 4
a_3 = a_2 + 3 + 4 + 4
a_4 = a_3 + 3 + 4 + 4 + 4
a_5 = a_4 + 3 + 4 + 4 + 4 + 4

Note that, in each step, the number of 4's being added is equal to the subscript on the a to the right of the equals sign. In other words,

a_2 = a_1 + 3 + 4*1
a_3 = a_2 + 3 + 4*2
a_4 = a_3 + 3 + 4*3
a_5 = a_4 + 3 + 4*4

We can generalize this to:

a_(n+1) = a_n + 3 + 4n

and I think that's what you're looking for.

Hope that helps!

Answer 4

Let use assume 3 to be the starting and increments by 3+(n-1)*4.

So,

a1 = 3 = 1*3+0*4 = 3
a2 = 3 + (3+4) = 2*3+1*4 = 10
a3 = 3 + (3+4) + (3+4+4) = 3*3 + 3*4 = 21
a4 = 3 + (3+4) + (3+4+4) + (3+4+4+4) = 4*3 + 6*4 = 36
a5 = 3 + (3+4) + (3+4+4) + (3+4+4+4) + (3+4+4+4+4) = 5*3 + 10*4

So in terms of 'n',
a(n) = a(n-1)+[3+(n-1)*4], and a(1) =3.

Hope this explains a lot

Answer 5

Second hexagonal numbers:
the n_th number is given as n(2n+1).

Answer 6

Yup. 2'nd hexagonal series.


Doug

Answer 7

the next one is 55.but i dont really have a way to explain it.