Math, asked by melfeenatilda, 1 month ago

if (.) represents the greatest integer function then evaluate the FOLLOWING sum[1/1]+[1/2]+[2/2]+[1/3]+[2/3]+[3/3]+[1/4]+[2/4]+[3/4]+[4/4]+[1/5]+.....upto the 2013th term​

Answers

Answered by amitnrw
0

Given : [1/1]+[1/2]+[2/2]+[1/3]+[2/3]+[3/3]+[1/4]+[2/4]+[3/4]+[4/4]+[1/5]+ upto the 2013th term​  

To Find : Value

Solution

[1/1]+[1/2]+[2/2]+[1/3]+[2/3]+[3/3]+[1/4]+[2/4]+[3/4]+[4/4]+[1/5]+ upto the 2013th term​  

only [1/1] , [2/2] , [3/3] , [4/4]  and so on will result in 1  , rest all will be 0

[1/1]  -  1 Term

[1/2]+[2/2] - 2 Terms

1/3]+[2/3]+[3/3] - 3 Term

Let say there are n such complete set

then sum = n

1 + 2 + 3 +                    + n  ≤ 2013

=> n(n + 1)/2 ≤ 2013

=> n(n + 1) ≤ 4026

63 * 64 = 4032 > 4026

62 * 63 = 3906 < 4026

Hence n = 62

can find Value of n using Quadratic equation formal

n² + n - 4026 ≤  0

n =( -1 ± √1 - 4(-4026) )/2  = (-1  ± 126.9)/2  =  62.95   ( ignoring - ve value )

62.95 < 63  Hence

Sum = 62  

[1/1]+[1/2]+[2/2]+[1/3]+[2/3]+[3/3]+[1/4]+[2/4]+[3/4]+[4/4]+[1/5]+ upto the 2013th term​   = 62

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