Found 3 solutions by ikleyn, greenestamps, Edwin McCravy:
Answer by ikleyn(52776) (Show Source): You can put this solution on YOUR website!
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In this problem, you are given a recurrent formula to calculate next term of the sequence
using its previous step. You also are given the first term.
So, to get the second term, simply substitute first term a1 = 3 into the formula
a2 = 2*3 + 5 = 6 + 5 = 11.
To get the third term, substitute the second term a2 = 11 into the formula
a3 = 2*11 + 5 =22 + 5 = 27.
Continue doing this way until you will get a5.
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!
The recursive formula says we get the next term in the sequence by doubling the current term and adding 5. Starting with 3 as the first term...
a(1) = 3
a(2) = 3(2)+5 = 11
a(3) = 11(2)+5 = 27
a(4) = 27(2)+5 = 59
a(5) = 59(2)+5 = 123
It would be very tedious to find the 50th term by continuing to use the recursive formula, so to find the 50th term we would like to find an explicit formula for the n-th term.
To do that, we could try the method of finite differences. Here are the first few terms of the sequence and the first and second differences:
3 11 27 59 123
8 16 32 64
8 16 32
That pattern of differences tells us that the formula for the n-th term is not a polynomial but instead is based on powers of 2.
Playing with the first few numbers in the sequence, we see that adding 5 to each term gives us a sequence of powers of 2:
8, 16, 32, 64, 128, ...
So...
a(1) is 8-5 = 2^3-5
a(2) is 16-5 = 2^4-5
a(3) is 32-5 = 2^5-5
...
And we see the pattern for the n-th term is
a(n) = 2^(n+2)-5
So the 50th term is
2^52-5 = 9007199254740987
My TI-84 calculator won't calculate that number; and I tried using excel but it fell one digit short of being able to calculate the exact number. I used an online program (pari) to find the exact answer.
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!
I wrote a program in LibertyBasic using the given recursion formula
and greenestamps' general formula. Here is the LibertyBasic progrem:
for n=1 to 50
if n=1 then a=3: goto 1
a=2*a+5
1 print n, a, 2^(n+2)-5
next
They are identical, as you can see from the complete output below. However, the
50th term does not agree with greenestamps' 50th term. He made a slight
calculator error and got the 51st term by calculating 2^53-5 instead of 2^52-5.
2a(n-1)-5
a(1)=3 a(n)=2n+2-5
1 3 3
2 11 11
3 27 27
4 59 59
5 123 123
6 251 251
7 507 507
8 1019 1019
9 2043 2043
10 4091 4091
11 8187 8187
12 16379 16379
13 32763 32763
14 65531 65531
15 131067 131067
16 262139 262139
17 524283 524283
18 1048571 1048571
19 2097147 2097147
20 4194299 4194299
21 8388603 8388603
22 16777211 16777211
23 33554427 33554427
24 67108859 67108859
25 134217723 134217723
26 268435451 268435451
27 536870907 536870907
28 1073741819 1073741819
29 2147483643 2147483643
30 4294967291 4294967291
31 8589934587 8589934587
32 17179869179 17179869179
33 34359738363 34359738363
34 68719476731 68719476731
35 137438953467 137438953467
36 274877906939 274877906939
37 549755813883 549755813883
38 1099511627771 1099511627771
39 2199023255547 2199023255547
40 4398046511099 4398046511099
41 8796093022203 8796093022203
42 17592186044411 17592186044411
43 35184372088827 35184372088827
44 70368744177659 70368744177659
45 140737488355323 140737488355323
46 281474976710651 281474976710651
47 562949953421307 562949953421307
48 1125899906842619 1125899906842619
49 2251799813685243 2251799813685243
50 4503599627370491 4503599627370491
Edwin
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