Let [α] denote the greatest integer ≤ α. Then [√1]+[√2]+[√3]+…+[√120] is equal to __________

Solution

Maths sir · 4 steps

Do not add 120 terms one by one. Group the n values that share the same integer part of their root.

1. Find the blocks. [√n] = k exactly when k² ≤ n ≤ (k+1)² - 1. That block has (k+1)² - k² = 2k+1 values of n, and each one contributes k.
2. Check the last block. For k = 10 the block runs from n = 100 to n = 120. That is exactly 21 = 2(10)+1 terms, so k runs over the full blocks from 1 to 10.

   Σₖ₌₁¹⁰ k(2k+1) = 2Σₖ₌₁¹⁰ k² + Σₖ₌₁¹⁰ k = 2(385) + 55

3. Use the formulas. Σ k² = (10 × 11 × 21)/6 = 385 and Σ k = (10 × 11)/2 = 55. So the sum is 770 + 55.
4. So. The required sum is 825.

The common slip: 615. Stopping at n = 99 and leaving out the block n = 100 to 120 gives 615, not 825. Those 21 terms each add 10.

The idea to keep. For sums of [√n], the value k repeats 2k+1 times, from k² to (k+1)² - 1. Always check whether the last block is full or cut short.

Answer from the NTA final answer key. Solution by PrepWiser.