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2 Answers
- ?Lv 711 months ago
The following is my work:
11,20,42,101,245,570,.....,T(n),...
.9,..22,.59,144,.325,.... D1
....13,.37,.85,.181,.....D2
......24,.48,.96,......D3
D3 can be rewritten as 24, 24*2,24*2^2,...
This results suggest a pattern that
T(n)=an^2+bn+c+d[2^(n-1)]
=>
a+b+c+d=11
4a+2b+c+2d=20
9a+3b+c+4d=42
16a+4b+c+8d=101
Solving the system, get
a=-5.5
b=1.5
c=-9
d=24
Thus, T(n)=24[2^(n-1)]-5.5n^2+1.5n-9.
Check:
11...20...42...101...245...570...1268...2723...5703...T(10),...
.. 9...22....59....144....325...698....1455...2980...D1
......13...37....85.....181...373...757.....1525....D2
..........24...48....96.....192...384....768.....D3
- Jeff AaronLv 711 months ago
The nth term of that sequence could be:
(1/5)n^5 - 2n^4 + 11n^3 - (55/2)n^2 + (383/10)n - 9