Question about Computers & Internet

The sum of all odd numbers, up to the odd number (2n-1) is

n^2. To illustrate this, think of the following example: What is

1 + 3 + 5 + ... + 997 + 999 ?

Well, 999 is of the form 2(500)-1, so n, in this case, is 500, so the

sum of all odd numbers (from 1) up to 999 is 500^2 or 250,00

So after placing formula you can get the answers easily

Thanks

Have a great day

Posted on Aug 29, 2010

For(i=1;i<n;i=i+2)

sum=sum+i;

n=no of terms,sum=0

Posted on Aug 29, 2010

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Posted on Jan 02, 2017

One or three of the numbers is odd. I'll leave the rest of your homework to you.

Mar 05, 2017 | Homework

Either one or three of the numbers are odd. I'll let you do the rest of your homework on your own.

Mar 05, 2017 | Homework

Mean= average =(sum of data values)/(number of data values)

Mode is the most frequent data value.

Median: Sort the data in ascending order. The element in the middle is the median.

If the number of data is odd, there will be one MIDDLE element. It is the median.

If the number of data is even, there will be two elements in the middle. Add the values of these middle elements and divide by 2. The result will be the median. It is not part of the data.

Mode is the most frequent data value.

Median: Sort the data in ascending order. The element in the middle is the median.

If the number of data is odd, there will be one MIDDLE element. It is the median.

If the number of data is even, there will be two elements in the middle. Add the values of these middle elements and divide by 2. The result will be the median. It is not part of the data.

Apr 22, 2014 | Texas Instruments TI-34II Explorer Plus...

Mean= average =(sum of data values)/(number of data values)

Mode is the most frequent data value.

Median: Sort the data in ascending order. The element in the middle is the median.

If the number of data is odd, there will be one MIDDLE element. It is the median.

If the number of data is even, there will be two elements in the middle. Add the values of these middle elements and divide by 2. The result will be the median. It is not part of the data.

Mode is the most frequent data value.

Median: Sort the data in ascending order. The element in the middle is the median.

If the number of data is odd, there will be one MIDDLE element. It is the median.

If the number of data is even, there will be two elements in the middle. Add the values of these middle elements and divide by 2. The result will be the median. It is not part of the data.

Apr 22, 2014 | Texas Instruments TI-34 Scientific...

If by "The sum of it's hundreds num is 4" you mean "The sum of its hundred number and ones number" the answer is 9173.

Jun 20, 2011 | Office Equipment & Supplies

No. Odd numbers less than 9 is 1, 3, 5, 7. Their sum is 1+3+5+7 = 16.

May 15, 2011 | Computers & Internet

this would be explained on the basis of two FACTS below.......

sum of two odd numbers are always even and

sum of to even numbers are also always even

then how 6 odd numbers (consider 3 pairs, 9 5 3 1 being odd numbers) can make an odd number 21..!!

SO SIMPLE

Make your basics strong...

sum of two odd numbers are always even and

sum of to even numbers are also always even

then how 6 odd numbers (consider 3 pairs, 9 5 3 1 being odd numbers) can make an odd number 21..!!

SO SIMPLE

Make your basics strong...

Aug 14, 2010 | Computers & Internet

Q.2# Dim n, sum as Double

Dim i as Integer

n=100

sum=0

for i=1 to n

sum=sum+1/n

next i

Q.3# Dim sum as Double

sum=0

for i=1 to 99

if i%2 <> 0 then sum=sum+i end if

next i

Dim i as Integer

n=100

sum=0

for i=1 to n

sum=sum+1/n

next i

Q.3# Dim sum as Double

sum=0

for i=1 to 99

if i%2 <> 0 then sum=sum+i end if

next i

Sep 16, 2008 | Microsoft AccRepair DS2 2004 for Fr...

#include <stdio.h>

int main()

{

int n1, n2, i;

int sum = 0;

n1 = 3;

n2 = 11;

for (i = n1; i <= n2; i++) {

if (i % 2 == 0) {

printf("Adding %d\n", i);

sum = sum + i;

}

}

printf("sum = %d\n", sum);

return 0;

}

int main()

{

int n1, n2, i;

int sum = 0;

n1 = 3;

n2 = 11;

for (i = n1; i <= n2; i++) {

if (i % 2 == 0) {

printf("Adding %d\n", i);

sum = sum + i;

}

}

printf("sum = %d\n", sum);

return 0;

}

Feb 21, 2008 | Computers & Internet

Aug 21, 2017 | Computers & Internet

Aug 21, 2017 | AVG Computers & Internet

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