It will become less steep.
Answer:
D) 18
Step-by-step explanation:
Isolate the variable by dividing each side by factors that don't contain the variable
Compare 1/7 to consecutive multiples of 1/9. This is easily done by converting the fractions to a common denominator of LCM(7, 9) = 63:
1/9 = 7/63
2/9 = 14/63
while
1/7 = 9/63
Then 1/7 falls between 1/9 and 2/9, so 1/7 = 1/9 plus some remainder. In particular,
1/7 = 1/9¹ + 2/63.
We do the same sort of comparison with the remainder 2/63 and multiples of 1/9² = 1/81. We have LCM(63, 9²) = 567, and
1/9² = 7/567
2/9² = 14/567
3/9² = 21/567
while
2/63 = 18/567
Then
2/63 = 2/9² + 4/567
so
1/7 = 1/9¹ + 2/9² + 4/567
Compare 4/567 with multiples of 1/9³ = 1/729. LCM(567, 9³) = 5103, and
1/9³ = 7/5103
2/9³ = 14/5103
3/9³ = 21/5103
4/9³ = 28/5103
5/9³ = 35/5103
6/9³ = 42/5103
while
4/567 = 36/5103
so that
4/567 = 5/9³ + 1/5103
and so
1/7 = 1/9¹ + 2/9² + 5/9³ + 1/5103
Next, LCM(5103, 9⁴) = 45927, and
1/9⁴ = 7/45927
2/9⁴ = 14/45927
while
1/5103 = 9/45927
Then
1/5103 = 1/9⁴ + 2/45927
so
1/7 = 1/9¹ + 2/9² + 5/9³ + 1/9⁴ + 2/45927
One last time: LCM(45927, 9⁵) = 413343, and
1/9⁵ = 7/413343
2/9⁵ = 14/413343
3/9⁵ = 21/413343
while
2/45927 = 18/413343
Then
2/45927 = 2/9⁵ + remainder
so
1/7 = 1/9¹ + 2/9² + 5/9³ + 1/9⁴ + 2/9⁵ + remainder
Then the base 9 expansion of 1/7 is
0.12512..._9
You can make it easier by replacing x^n with another variable, factoring, then putting x^n back in the end.
Using exponent and algebra rules, rewrite x^2n - 2x^n + 1 as
(x^n)^2 - (2 x x^2) + 1
Then, let x^n = m.
m^2 - 2m + 1
Now factor that: (m - 1)^2
And now put x^n back: (x^n - 1)^2
