Question 60460
I hope I can help. You'll do yourself (and your grades) a big favor
if you get in the habit of always checking your answers. Simply put,
that just means plugging your solution back into the original problem
and seeing if it's really true. Of course, you need to arrive at a solution
in the first place.
(1) 1,2,3,4,5,6
This has lots of names: counting sequence, cardinal numbers, etc.
Let's say {{{k[n]}}}= the nth term, then
{{{k[n] = n}}} Remember what I said. Test it, even if it appears simple.
{{{k[1] = 1}}}
{{{k[2] = 2}}}
{{{k[3] = 3}}}
looks OK
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(2) 1,2,4,8,16
Don't rush to an answer. What do you see? Every term is even but the first
Ok, if they're even, they're divisible by 2. Write them differently
1
2
2*2
2*2*2
2*2*2*2
This is called a power sequence. You just have to write it with exponents.
n=1.....{{{2^0}}}
n=2.....{{{2^1}}}
n=3.....{{{2^2}}}
n=4.....{{{2^3}}}
n=5.....{{{2^4}}} etc.
Compare the exponent with it's position in the sequence. It looks like,
if the position is n, then the exponent is n-1, so if k is the general
term,
{{{k[n] = 2^(n-1)}}} As always, check the answer
{{{k[1] = 2^0}}}
{{{k[2] = 2^1}}}
{{{K[3] = 2^2}}} ETC.
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#10. 1,-1,1,-1,1,-1,...
Again, k is the general term
{{{k[n] = (-1)^(n+1)}}}
check
{{{k[1] = (-1)^(1+1)}}}
{{{k[2] = (-1)^(2+1)}}}
{{{k[3] = (-1)^(3+1)}}}
These look OK. With practice you'll spot these things
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That's all I have time for. Hope it helps.