所有数字的总和

Question

I need to write a function that calculates the sum of all numbers n.

Row 1:          1 
Row 2:         2 3 
Row 3:        4 5 6 
Row 4:       7 8 9 10 
Row 5:     11 12 13 14 15 
Row 6:   16 17 18 19 20 21 

It helps to imagine the above rows as a 'number triangle.' The function should take a number, n, which denotes how many numbers as well as which row to use. Row 5's sum is 65. How would I get my function to do this computation for any n-value?

For clarity's sake, this is not homework. It was on a recent midterm and needless to say, I was stumped.

Answer 1

The leftmost number in column 5 is 11 = (4+3+2+1)+1 which is sum(range(5))+1 . This is generally true for any n .

So:

def triangle_sum(n):
    start = sum(range(n))+1
    return sum(range(start,start+n))

---

As noted by a bunch of people, you can express sum(range(n)) analytically as n*(n-1)//2 so this could be done even slightly more elegantly by:

def triangle_sum(n):
    start = n*(n-1)//2+1
    return sum(range(start,start+n))

Answer 2

A solution that uses an equation, but its a bit of work to arrive at that equation.

def sumRow(n):
    return (n**3+n)/2

Answer 3

The numbers 1, 3, 6, 10, etc. are called triangle numbers and have a definite progression. Simply calculate the two bounding triangle numbers, use range() to get the numbers in the appropriate row from both triangle numbers, and sum() them.

Answer 4

def sum_row(n):
    final = n*(n+1)/2
    start = final - n
    return final*(final+1)/2 - start*(start+1)/2

or maybe

def sum_row(n):
    final = n*(n+1)/2
    return sum((final - i) for i in range(n))

How does it work:

The first thing that the function does is to calculate the last number in each row. For n = 5, it returns 15. Why does it work? Because each row you increment the number on the right by the number of the row; at first you have 1; then 1+2 = 3; then 3+3=6; then 6+4=10, ecc. This impy that you are simply computing 1 + 2 + 3 + .. + n, which is equal to n(n+1)/2 for a famous formula.

then you can sum the numbers from final to final - n + 1 (a simple for loop will work, or maybe fancy stuff like list comprehension) Or sum all the numbers from 1 to final and then subtract the sum of the numbers from 1 to final - n, like I did in the formula shown; you can do better with some mathematical operations

Answer 5

Here is a generic solution:

start=1
n=5
for i in range(n):
    start += len (range(i))
answer=sum(range(start,start+n))

As a function:

def trio(n):
    start=1
    for i in range(n):
            start += len (range(i))
    answer=sum(range(start,start+n))
    return answer

Answer 6

def compute(n):
   first = n * (n - 1) / 2 + 1
   last = first + n - 1
   return sum(xrange(first, last + 1))