# Find the sum of all the numbers in the following array:

1 2 3 4 5 6........100
2 4 6 8 10 12 ......200
3 6 9 12 15 18 ......300
.
.
.
100 200 300 400 500 600 . . . 100(x)100

Can anyone help me??

I'll suggest how to solve this, but you'll need to do the calculating yourself.
Observe that the first row is:
1+2+3+...+99+100=row_one
where row_one is the sum of row one.
If you know how to sum this you'll have no problem doing the rest of the array.
The second row is:
2+4+6+8+...+198+200=2(1+2+3+4+...+99+100)=
2*row_one
The third row is 3+6+9+...+297+300=3(1+2+3+4...+99+100)=
3*row_one
The sum of the array is then:
row_one +
2*row_one +
3*row_one +

.
.
.
k*row_one (the k-th row) +
.
.
.
100*row_one
Thus the sum of the array is
1*row_one+2*row_one+3*row_one+...100*row_one=
row_one(1+2+3+...+99+100)=row_one*row_one
or row_one^2
Now what is row_one?

Roger, Thank you! I did solve it like that, but it did not look right! I e-mailed my teacher to ask her but, she has not e-mailed me back yet. It is a homework problem for tomorrow. Thanks for letting me know I was right! I am not very good in math! LOL!

x<-2>3=

anybody cam help this problem.......this is my assignment to be pass tomorrow....

pls..pls...help me....

Andy, I'm not exactly sure what your question is. The statement doesn't make sense as it stands.
x<-2>3=
Can you clarify what you mean here?
Also, you should start a new thread.
Do you see the Post a New Question link above?

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