Mathematics question
2007-07-23 3:53 am
If the general term of a sequence is given by T(n) = n^2-n. If T(k) = 30, k=?
回答 (2)
2007-07-23 3:58 am
✔ 最佳答案
T(k) = 30k^2 - k = 30
k^2 - k - 30 = 0
(k - 6)(k + 5) = 0
k = 6 or k = -5
Since k is positive, k = 6
2007-07-23 8:59 am
Since T(n) = n^2-n
and T(k) = 30
so T(k)=k^2-k=30
k^2-k-30=0
(k-6)(k+5)=0
k=6 or -5
checking:
sub k=6 in to k^2-k=30
L.H.S=(6)^2-6
=36-6
=30=R.H.S
so k=6 is a solution of the equation
sub k=-5 into k^2-k=30
L.H.S=(-5)^2-(-5)
=25+5
=30=R.H.S
so the solution of the equation is k=6 or -5
and T(k) = 30
so T(k)=k^2-k=30
k^2-k-30=0
(k-6)(k+5)=0
k=6 or -5
checking:
sub k=6 in to k^2-k=30
L.H.S=(6)^2-6
=36-6
=30=R.H.S
so k=6 is a solution of the equation
sub k=-5 into k^2-k=30
L.H.S=(-5)^2-(-5)
=25+5
=30=R.H.S
so the solution of the equation is k=6 or -5
參考: My brain
收錄日期: 2021-04-13 00:50:37
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