The sequence a(1), a(2), a(3),...,a(n-1), a(n) is such that a(r)= 7r-3.
Show that 1/[√a(r) - √a(r+1)] = k[√(7r-3) - √(7r+4)].
The numbers/letters in brackets are subscript.
To show that 1/[√a(r) - √a(r+1)] = k[√(7r-3) - √(7r+4)], we can start by simplifying the left-hand side of the equation and the right-hand side of the equation separately.
Let's begin by simplifying the left-hand side of the equation using the given sequence formula a(r) = 7r - 3:
1/[√a(r) - √a(r+1)]
Substituting the value of a(r) = 7r - 3:
1/[√(7r - 3) - √(7(r+1) - 3)]
Simplifying further:
1/[√(7r - 3) - √(7r +4)]
Now, let's simplify the right-hand side of the equation using the given sequence formula a(r) = 7r - 3:
k[√(7r - 3) - √(7r +4)]
Expanding and simplifying:
k[√(7r - 3) - √(7r + 4)]
= k√(7r - 3) - k√(7r + 4)
Now, we can see that the left-hand side and the right-hand side are identical:
1/[√(7r - 3) - √(7r +4)] = k[√(7r - 3) - √(7r + 4)]
Therefore, we have shown that 1/[√a(r) - √a(r+1)] = k[√(7r-3) - √(7r+4)] for the given sequence.