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 need to manipulate the left-hand side (LHS) expression until it matches the right-hand side (RHS) expression. Let's start by evaluating the LHS expression using the given sequence.
Given that a(r) = 7r-3, we can rewrite the LHS expression as follows:
1/[√(7r-3) - √(7(r+1)-3)]
Next, let's manipulate the expression inside the square roots to simplify it further:
√(7r-3) - √(7(r+1)-3)
= √(7r-3) - √(7r+7-3)
= √(7r-3) - √(7r+4)
Now, let's substitute this simplified expression back into the LHS expression:
1/[√(7r-3) - √(7r+4)]
Comparing this with the RHS expression k[√(7r-3) - √(7r+4)], we can see that the expressions are identical.
Therefore, we have shown that 1/[√a(r) - √a(r+1)] = k[√(7r-3) - √(7r+4)].