Question
Quantitative Aptitude Question on Basics of Numbers
Let an be the largest integer not exceeding n. Then the value of a1+a2+⋯+a50 is
We are asked to find the sum of a1+a2+⋯+a50, where an=⌊n⌋.
The value of an is the greatest integer less than or equal to n. To find the sum, we can break the sum into intervals where ⌊n⌋ remains constant. The value of ⌊n⌋ will stay constant for values of n within certain intervals.
- For n=1 to 3, ⌊n⌋=1 (3 terms).
- For n=4 to 8, ⌊n⌋=2 (5 terms).
- For n=9 to 15, ⌊n⌋=3 (7 terms).
- For n=16 to 24, ⌊n⌋=4 (9 terms).
- For n=25 to 35, ⌊n⌋=5 (11 terms).
- For n=36 to 48, ⌊n⌋=6 (13 terms).
- For n=49 and 50, ⌊n⌋=7 (2 terms).
Now, calculate the total sum:
Total sum =1×3+2×5+3×7+4×9+5×11+6×13+7×2
=3+10+21+36+55+78+14=217.
Thus, the value of a1+a2+⋯+a50=217.
Solution
We are asked to find the sum of a1+a2+⋯+a50, where an=⌊n⌋.
The value of an is the greatest integer less than or equal to n. To find the sum, we can break the sum into intervals where ⌊n⌋ remains constant. The value of ⌊n⌋ will stay constant for values of n within certain intervals.
- For n=1 to 3, ⌊n⌋=1 (3 terms).
- For n=4 to 8, ⌊n⌋=2 (5 terms).
- For n=9 to 15, ⌊n⌋=3 (7 terms).
- For n=16 to 24, ⌊n⌋=4 (9 terms).
- For n=25 to 35, ⌊n⌋=5 (11 terms).
- For n=36 to 48, ⌊n⌋=6 (13 terms).
- For n=49 and 50, ⌊n⌋=7 (2 terms).
Now, calculate the total sum:
Total sum =1×3+2×5+3×7+4×9+5×11+6×13+7×2
=3+10+21+36+55+78+14=217.
Thus, the value of a1+a2+⋯+a50=217.