2007-05-25 03:30:20 · 6 answers · asked by galin 2 in Science & Mathematics ➔ Mathematics
The sum of numbers from a to b,where a>b is given by-
{a*(a+1)-b*(b-1)} /2
So, sum of digits from 3432 to 223232321323
={(223232321323 *223232321324) - (3432*3431)} /2
=(49832669283478352791652 - 11775192)/2
=49832669283478341016460 /2
=24916334641739170508230
2007-05-25 08:38:32 · answer #1 · answered by Happy 3 · 0⤊ 0⤋
Take the average of the end points and multiply by the number of numbers.
(223232321323+3432)/2 * (223232321323-3432+1)
= 2.49163346... Ã 10^22
2007-05-25 10:35:39 · answer #2 · answered by Blank 2 · 0⤊ 0⤋
Ok first number is (a) = 3432
Last number is (l) = 223232321323
The series is 3432,3433,3434,............,223232321323
Therefore it is in the form of arithmetic progression(AP) with common difference "d"=1
Therefore sum of the numbers = (n/2)(a+l)
=(223232321323-3432)*(3432+223232321323)/2
2007-05-27 13:57:59 · answer #3 · answered by sriram t 3 · 0⤊ 0⤋
Are you serious? There are 3 steps.
Find the sum from 1 to 3432 = (3433*3232/2)
Find the sum from 1 to the other number
Subtract first sum from second.
2007-05-25 10:38:21 · answer #4 · answered by cattbarf 7 · 2⤊ 1⤋
249 1633 4641 7391 7050 4798
Value of 3432 is not included in answer and it is included in a value of sum of numbers 1...3432!
With best regards!
2007-05-29 05:49:23 · answer #5 · answered by kkr 3 · 0⤊ 0⤋
The sum of integers, 1 through n, is: n(n + 1) / 2. As cat... suggested, subtract the smaller sum from the larger.
2007-05-25 10:42:25 · answer #6 · answered by Darlene 4 · 0⤊ 0⤋