1,5,14,30,55
then 1,5,17,53 and 161
2007-06-20 17:59:23 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1,5,14,30,55
1 ---> 5 = +4 = 2²
5 ---> 14 = +9 = 3²
14 ---> 30 = +16 = 4²
30 ---> 55 = +25 = 5²
next number, xx
55 ---> xx = 55 + 6² = 91
1,5,17,53 and 161
1 ---> 5 = +4
5 ---> 17 = +12 = 12/4 = 3
17 ---> 53 = +36 = 36/12 = 3
53 ---> 161 = +108 = 108/36 = 3
next number, yy
161 ---> yy = 3*108 + 161 = 485
2007-06-20 21:15:58 · answer #1 · answered by jurassicko 4 · 0⤊ 0⤋
1, 5, 14, 30, 55
1 = 1^2
5 = 1^2+2^2
14 = 1^2+2^2+3^2
30 = 1^2+2^2+3^2+4^2
55 = 1^2+2^2+3^2+4^2+5^2
sum of squares
for n=1 to 5 : sum n^2
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1, 5, 17, 53, 161
1 = 2(3^0)-3^0
5 = 2(3^0+3^1)-3^1
17 = 2(3^0+3^1+3^2)-3^2
53 = 2(3^0+3^1+3^2+3^3)-3^3
161 = 2(3^0+3^1+3^2+3^3+3^4)-3^4
2 * (sum of powers of 3) - 3^(last power)
for n=0 to 4 : 2*(sum 3^n)-3^n
2007-06-21 02:46:01 · answer #2 · answered by jimschem 4 · 0⤊ 0⤋
In the first sequence Tn+1 = Tn + (n+1)^2
In the second sequence Tn+1 = 4Tn - 3 Tn-1,
2007-06-21 01:09:28 · answer #3 · answered by cattbarf 7 · 0⤊ 0⤋
I'm not sure about the first sequence, but the second one is nx3=2
2007-06-21 01:09:25 · answer #4 · answered by volleyball girl 2 · 0⤊ 1⤋
1,5,14,30,55,91
1,5,17,53,161,485
2007-06-21 01:15:57 · answer #5 · answered by jesem47 3 · 0⤊ 0⤋