Final answer:
To prove that a₁^(n+1) = a^b, we can use the formula for the nth term of a geometric sequence. By simplifying the expression and using the property of exponents, we conclude that a₁^(n+1) is equal to a^b.
Step-by-step explanation:
To prove that a₁^(n+1) = a^b, we need to use the formula for the geometric mean. The formula for finding the nth term of a geometric sequence is a_n = a * r^(n-1), where a is the first term and r is the common ratio. In this case, a₁ is the first of n geometric means between a and b. So we have:
a₁ = a * r^(1-1) = a * r^0 = a
Now, we can use the formula for the nth term:
a_n = a * r^(n-1)
a₁^(n+1) = a * r^(n-1)^(n+1)
To simplify, we use the property that a^b * a^c = a^(b+c):
= a * r^(n*n-1)
= a * r^(n^2 - n)
= a * r^(n^2) * r^(-n)
Using the property r^(-n) = 1 / r^n, we get:
= a * r^(n^2) / r^(n)
= a * (r^(n^2) / r^n)
Since a_n = a * r^(n-1), we can rewrite r^(n^2) / r^n as a_n+1 / a_n:
= a * (a_n+1 / a_n)
= a * (a * r^n / a * r^(n-1))
= a * (a * r^n / a * r^(n-1))
= a * (a * r^(n-n+1) / a * r^(n-1))
= a * (a * r / a * 1)
= a * r
= a^b