Q13/048) Without using a calculator find which is greater √1001 + √+999 or 2 √1000

We have √ (n+1) - √n = 1/(√(n+1) + √n)…  (1)

And √(n) - √(n-1)  = 1/(√(n + √(n-1)) …2

As

1/(√(n+1) + √n) < 1/(√(n + √(n-1)) ( as denominator of LHS is larger)

So √ (n+1) - √n < √(n) - √(n-1)

So √ (n+1) +  √(n-1)  < 2√(n)

Putting n = 1000 we get

√1001 + √+999 < 2 √1000

or 2 √1000 is larger