- #1
silina01
- 12
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Define the sequence of integers a1, a2, a3,... as follows:
a1 = 3
a2 = 6
an = 5an-1 - 6an-2 + 2 for all n ≥ 3
Prove that an = 1 + 2n-1 + 3n-1
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my attempt:
base case:
n=1
1+ 20 +30
= 1
= a1
therefore n=1 holds
n=2
1+ 21 +31
= 6
= a2
therefore n=2 holds
Assume k holds for all k > n, n≥3 then
an = 5an-1 - 6an-2 + 2
= 5(1+2n-2 + 3n-2) -6(1 + 2n-3+ 3n-3) +2
= 1 + 10n-2 + 15n-2 -12n-3 -18n-3
I have no idea what to do from here.
a1 = 3
a2 = 6
an = 5an-1 - 6an-2 + 2 for all n ≥ 3
Prove that an = 1 + 2n-1 + 3n-1
------------------------------------------------------------------------------------------------
my attempt:
base case:
n=1
1+ 20 +30
= 1
= a1
therefore n=1 holds
n=2
1+ 21 +31
= 6
= a2
therefore n=2 holds
Assume k holds for all k > n, n≥3 then
an = 5an-1 - 6an-2 + 2
= 5(1+2n-2 + 3n-2) -6(1 + 2n-3+ 3n-3) +2
= 1 + 10n-2 + 15n-2 -12n-3 -18n-3
I have no idea what to do from here.
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