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Let [x] denote the greatest integer less than or equal to x. Find the value of [2/5]+[4/5]+[6/5]+...+[48/5]?

6 Answers

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  • 1 decade ago
    Favourite answer

    Greetings,

    We must group the values by multiples of 5

    2,4 < 5 so 2*0

    6, 8, < 10 so 2*1

    10, 12, 14 < 15 so 3*2

    16, 18 < 20 so 2*3

    20, 22, 24< 25 so 3*4

    26, 28 < 30 so 2*5

    30, 32, 34 < 35 so 3*6

    36, 38 < 40 so 2*7

    40, 42, 44 < 45 so 3*8

    46, 48 < 50 so 2*9

    2(1 + 3 + 5 + 7 + 9) + 3(2 + 4 + 6 + 8)

    = 2*25 + 3*20

    = 50 + 60

    = 110

    Regards

  • 1 decade ago

    If you're interested in just the answer, a program can quickly calculate this for you. If you're interested in how to obtain this answer, I suggest you review arithmethetic series.

    --- Program Begin ---

    0 + 0 + 1 + 1 + 2 + 2 + 2 + 3 + 3 + 4 + 4 + 4 + 5 + 5 + 6 + 6 + 6 + 7 + 7 + 8 + 8 + 8 + 9 + 9

    sum = 110

    --- Program End ---

    Source(s): my programming skills: public class NumberTest { public static void main(String[] args) { int sum = 0; System.out.println("--- Program Begin ---"); for (int i=2; i<=48; i = i+2) { sum += f(i/5); System.out.print( f(i/5) + " + " ); } System.out.println("\n\nsum = " + sum + "\n\n" ); System.out.println("--- Program End ---"); } public static int f(double x) { int k = (int)Math.floor(x); return k; } }
  • 1 decade ago

    Notice that:

    2/5 + 48/5 = 50/5 = 10

    4/5 + 46/5 = 50/5 = 10

    Get the idea?

    I think there are 12 pairs like this, so 12*10 = 120

  • 1 decade ago

    You can pair these up with greatest and least which are 2/5 + 48/5 = 10, and then the next pair

    4/5 + 46/5 = 10, and keep going.

  • 1 decade ago

    apply the same concept that is applied to find sum of consecutive N natural numbers.

    note [4/5]=0, [48/5]=9 etc. etc.

    assume the sum is S. write it forward(as presented) and in reverse order.

    S=[ 2/5]+[ 4/5]+[ 6/5]..............+[44/5]+[46/5]+[48/5]

    S=[48/5]+[46/5]+[44/5]...........+[ 6/5] +[ 4/5] + [2/]

    summing the two series

    2S=(0+9)=(0+9)+(1+8) +(8+1)+(9+0)+(9+0)

    There are 24 items of summation in the series.

    2S=24*9

    S=12*9=108

  • ?
    Lv 4
    5 years ago

    often, [-x] + [x] = -a million if x isn't an integer and 0 if x is an integer. WLOG, we can assume that x is helpful. If x isn't an integer, then there exists an integer n such that n < x < n + a million wherein case, [x] = n. We be conscious that -n - a million < -x < -n. wherein case [-x] = -n - a million. hence [x] + [-x] = -a million whilst x is an integer, then [x] + [-x] = x + -x = 0.

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