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secondary 4 | E Maths
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Anyone can contribute an answer, even non-tutors.
Pls help with all parts thanks!!
Date Posted: 5 years ago
Views: 379
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Jiayous (-: and soz about part b and e. my brain a bit tired ah. paiseh. maybe someone else can help with it (:
Date Posted:
5 years ago
Its ok, thanks for the other parts too!! :)
you’re welcome (-:
(b)
from part (c),
Tn = n² + 5n + 3
for all even n's, n² is even, 5n is even,
and even plus even is even,
hence, Tn = n² + 5n + 3 is odd
since even plus odd is odd.
for all odd n's, n² is odd, 5n is odd,
and odd plus odd is even,
hence, Tn = n² + 5n + 3 is odd
since even plus odd is odd.
(e)
Tn = n² + 5n + 3
and the consecutive term,
T(n+1) = (n+1)² + 5(n+1) + 3
= n² + 2n + 1 + 5n + 5 + 3
= n² + 7n + 9
so their difference is
n² + 7n + 9 - (n² + 5n + 3)
= 2n + 6
for all n values,
(2n+6) is greater than 4,
hence their difference cannot be 4.
hope this helps.
from part (c),
Tn = n² + 5n + 3
for all even n's, n² is even, 5n is even,
and even plus even is even,
hence, Tn = n² + 5n + 3 is odd
since even plus odd is odd.
for all odd n's, n² is odd, 5n is odd,
and odd plus odd is even,
hence, Tn = n² + 5n + 3 is odd
since even plus odd is odd.
(e)
Tn = n² + 5n + 3
and the consecutive term,
T(n+1) = (n+1)² + 5(n+1) + 3
= n² + 2n + 1 + 5n + 5 + 3
= n² + 7n + 9
so their difference is
n² + 7n + 9 - (n² + 5n + 3)
= 2n + 6
for all n values,
(2n+6) is greater than 4,
hence their difference cannot be 4.
hope this helps.
Thank you!!
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