Video Transcript
Find, in terms of 𝑛, the general
term of the sequence one-fourth, nine over 16, 81 over 64, 729 over 256, and so
on.
In this question, we’re given the
first four terms of this sequence. There doesn’t look like there’s a
common difference between the terms, so we could say that this is definitely not an
arithmetic sequence. We can check if it’s a geometric
sequence which would have a common ratio between the terms by seeing if we can work
out what that common ratio would be.
In order to find the ratio 𝑟
between the first two terms, we would take the second term, nine over 16, and divide
by the term before it, one-quarter. We can recall that to divide by
one-quarter, that would be equivalent to multiplying by the reciprocal, which would
be four over one. We can simplify the four on the
numerator and the 16 on the denominator by taking out a factor of four. We then multiply our numerators and
denominators. Nine times one would give us nine,
and four times one gives us four. We can check if there’s the same
ratio between the third term and the second term. So, we calculate 81 over 64 divided
by nine over 16. We can see that the reciprocal of
nine over 16 would be 16 over nine. And we simplify the fractions
before multiplying the numerators and denominators, which gives us the same common
ratio 𝑟 of nine over four. And if we check this by working out
what we need to multiply the third term 81 over 64 by in order to get the fourth
term of 729 over 256, we get the same answer of nine over four.
So, let’s think about how we would
find the general term of this sequence. We can recall that the general term
is another way of saying the 𝑛th term. We can recall that the formula to
find the 𝑛th term 𝑎 sub 𝑛 is that 𝑎 sub 𝑛 is equal to 𝑎 times 𝑟 to the power
of 𝑛 minus one. The 𝑟-value represents the common
ratio, and the 𝑎-value represents the first term of the sequence. We’ve already established that 𝑟
is equal to nine over four, and the 𝑎-value, the first term in the sequence, would
be one-quarter. We can then take the values of 𝑎
and 𝑟 and plug them into this formula. This gives us 𝑎 sub 𝑛 equals
one-quarter times nine-quarters to the power of 𝑛 minus one.
While this is a perfectly valid
answer for the 𝑛th term of the sequence, it might be worth seeing if we can
simplify this right-hand side any further. We can start our simplification by
thinking about what happens when we have a power of a fraction. A fraction with a certain exponent
value is equivalent to the numerator with that exponent over the denominator with
that exponent. If we then consider these fractions
multiplied, multiplying the numerators one times nine to the power of 𝑛 minus one
would give us the value on the numerator of nine to the power of 𝑛 minus one. Multiplying the denominators would
give us four times four to the power of 𝑛 minus one.
If we look at the denominator, we
can apply another exponent rule that 𝑥 to the power of 𝑎 multiplied by 𝑥 to the
power of 𝑏 is equal to 𝑥 to the power of 𝑎 plus 𝑏. On the denominator, the 𝑥-value
here would be four. And our 𝑎- and 𝑏-values, that’s
the exponents, would be represented by one and 𝑛 minus one. Adding one and 𝑛 minus one would
leave us with the value of 𝑛. So, we’ve found a fully simplified
answer for the general term of this sequence in terms of 𝑛. It’s nine to the power of 𝑛 minus
one over four to the power of 𝑛.
