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Find a formula for the nth general term using powers of n?
Please help me with this maths question...
Find a formula for the nth general term using powers of n, factoral expressions involving n, and/or a constant to the nth power, starting with n=1. The first four terms are (n = 1,2,3,4):
2, 2*4, 2*4*6*8*10, 2*4*6*8*10*12*14, ...
Thank you
oops - yes, the second term was meant to be 2*4*6 :)
so, should be...
2,
2*4*6,
2*4*6*8*10,
2*4*6*8*10*12*14
thank you
4 Answers
- Anonymous1 decade agoFavourite answer
The first four terms are
1! * (2^1)
2! * (2^2)
5! * (2^5)
7! * (2^7)
It is clear that the general term is going to be something like x! * (2^x) but the formula for x is not easy to see. Can you add two or three more terms?
If the second term had been 2*4*6 then x would have just been the odd integers and
u(n) = (2n - 1)! * (2^(2n - 1)).
or even simpler would have been
u(n) = (4n - 2)!!
Source(s): Also a retired mathematics teacher. - HemantLv 71 decade ago
Second term should be 2.4.6.
Then, as the Mathsman correctly observed,
the n th term will be
( 2n - 1 )!. 2^ (2n-1).
- kincerLv 44 years ago
the 1st series, a million,9,17,25,... the n th term is 8(n-a million) + a million then the 1st term is, 8(a million-a million) + a million =a million the 2d term = 8(2-a million) +a million = 9, and so on. the subsequent one is obviously a chain of squares, n th term is (n+a million)^2
