ii) 2+10+30+68+130....
2007-02-24 02:50:27 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
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2007-02-24 02:57:54 · answer #1 · answered by Anonymous · 2⤊ 1⤋
Answer to the first:
( [ 3^(n+1) - 3 ] / 4 ) - n/2
Check:
If n = 1, the answer should be 1
3^(2) = 9
9-3 = 6
6 / 4 = 1.5
1.5 - 1/2 = 1
Works out
If n = 2, the answer should be 5 (1+4)
3^(3) = 27
9 - 3 = 24
24 / 4 = 6
6 - 2/2 = 5
Works out
Second is broken down into the pattern:
[1*2] + [2*2 + 1*6] + [3*2 + 4*6] + [4*2 + 10*6] + [5*2 +20*6]...
But I don't have the time to right the summation formula right now. Will do so later tonight if someone else hasn't already.
2007-02-24 03:25:36 · answer #2 · answered by Anonymous · 0⤊ 0⤋
this person wants a general solution...
there is no common difference, but the first is a summation of 3^n
2007-02-24 03:03:41 · answer #3 · answered by creepy_mitch 2 · 0⤊ 0⤋
a(nth number) = a1 + (n-1)d where a1 is the 1st term and d is the common difference
use that formula to solve
cheers
2007-02-24 02:58:02 · answer #4 · answered by Sum Girl 4 · 0⤊ 0⤋
(3power n -1)/2
2007-02-24 04:49:34 · answer #5 · answered by ravendra s 1 · 0⤊ 0⤋
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2007-02-25 23:00:43 · answer #6 · answered by fearlessme 2 · 0⤊ 0⤋