c++ program to find the sum of a series?
920270
Here's a start.\n\nI see you are using 'getch()' from conio.h\n\nI assume this is to insert a pause into the program?\n\nThis is not an ideal bit of coding as it is no longer supported by all compilers. Do you know that your teacher wants you to use it? (I assume this is homework?)\n\nHere's some discussion about it if you're interested\nhttp://forums.devarticles.com/c-c-help-52/c-problems-with-getch-6229.html\n\nIf you make an attempt to write some code we may be able to help you more.\n\n\n# include<iostream>\nusing namespace std;\n\nint main()\n{\n   \n  getch();\n\n   return 0;\n}\n\nNOTE: Be careful about using the answers supplied below.\nmerven uses 'void main()' which is not standard c++.\nthe other example uses some C rather than C++ in places.\n\nAnd this is from smart people who know what they are talking about and are trying to help you. No disrespect of any kind intended from me towards them. Just that what they are coding may not meet your teacher's approval (depending on her/his preferences).\n\nAre you prepared to do any work at all to help yourself and therefore help us to help you?\n\n///
729280
you can sum a row or column of cells by typing =sum(a1:a5)\n\nReplacing a1 and a5 with appropriate cells.  you can also use this with difference or average in place of sum
588997
I use "C++ How to Program" by Deitel & Deitel.\nIt's really good
1054159
First, get rid of the brackets (multiply the terms inside the brackets by -1):\n\n7-2x^2 +2x -2 -2x -5 +2x^2\n\nCollect like terms:\n\n2x^2 -2x^2 +2x -2x +7-2-5 = 0\n\nIt's easy to see the sum is zero.
943372
u have to ponit it or use\n\nthese are the 2 way's:\n\n#include <iostream>\nusing namespcae std;\nint main()\n{\nint num[8];\nint index;\n\nnum[0] = 42;\nnum[1] = 2001;\nnum[2] = 7;\nnum[3] = 180;\nnum[4] = 99;\nnum[5] = 1993;\n\nfor(index = 0; index <= 5; index++)\n{\nstd::cout<< index[num] << endl;\n}\n system("pause");\nreturn 0;\n}\n\nor\n\n#include <iostream>\n#define MAX 20\nusing namespcae std;\n\nint main()\n{\n\nint a[MAX] = {23,43,4,5,34,343,12,14};\nint b;\n\ncout.setf(ios::right);\n\nfor(b = 0; b < MAX; b++){\n\ncout << endl << endl;\n\ncout << "The Elemnt in ";\n\ncout.width(2);\n\ncout << (b +1) << " is ";\n\ncout.width(2);\n\nstd::cout << b[a];\n}\nreturn 0;\n}
1092372
6\n1+(1*2)+3\n1+2+3\n3+3\n6\n\nuse order of opperations.  PEMDAS
294753
Write 9/[n(n-1)]=9/(n-1)-9/n. The sum is then a telescoping series, so the sum from n=2 to n=9 will be\n9/1-9/2+2/2-9/3....+9/8-9/9=9-1=8.\n\nThe nth partial sum will be 9-9/n, which has a limit\nof 9 as n-->infty. Thus, the sum of the infinite series is 9.
1115229
Just add them up using the calculator:\n1000 + 5999 + 9888 = 16887\n\nOr if you don't have a calculator:\n*111\n---------\n**1000\n+5999\n+9888\n=====\n16887\n=====
34052
I do not know a preprogrammed utility to do this, but can point you in the direction that you might need to go.  See the reference below for the math and an on-line converter:
457195
the simple program given above if fine.. in terms of the conncept.. but there's a bug... \nthe problem is with looping.. coz when u enter a number such as 80301  or 55302 ( a no. having 0 in between), the loop breaks at the point when zero is encountered.\n\n so the output in the first eg. is  8 0 ,sum= 8 (instead of correct 8 0 3 0 1 , sum=12\n\n\nso the logic of handling the number is absolutely correct but the looping is a comparatively difficult job in this case!
681952
Here you go:\nhttp://en.wikipedia.org/wiki/Fibonacci_series
870473
Short answer, no. Inside your loop you keep resetting the result variable back to x. You won't get what you need.\n\nPlus, you need a main function if this program is going to compile, right? Use cin to get the x and n variables from the user and cout to output the result. Also, don't forget\n\nusing namespace std;\n\nBut as for the actual power function, it should look something like this.\n\ndouble PowerOf(double x, int n)\n{\n double result = 1;\n for (int i = 0; i < n; i++)\n {\n  result *= x;\n }\n return result;\n}\n\nHowever, I believe if you #include <cmath>   you can just use the pow function to do all the work for you.
964471
Hi there you can go to www.cplusplus.com\nOr if you want help ina specific topic just goolgle iT. fOR example\narrays in C++ go to i'm feeling lucky
5645
Yes, it does. \n\nTo be able to see the sum, right click on the status bar. You will be offered a choice of what you can see there - Average, Count, Min, Max, Sum etc. Choose Sum.\n\nThen select the cells which contain numbers (as in previous numbers). The sum will be displayed in the status bar.
175327
Try this, maybe?\n\nhttp://mathforum.org/library/drmath/drmath.elem.html
283059
no sorry thats not a valid equation\nthe second one is wrong \nthe corrected eqation will be\n[200+.........+300=5*1050]\n\nbut lol \nsince [100+105......+200]=3*1050\n[300+.......400]=7*1050\n[400+......+500]=9*1050\nand so it goes on\nCONGRAGULATION!!!!!!!\nyou just helped in finding a new series
201475
Do the math or at least show us the factors to get the sum and then we can decide whether or not it's true.  You can't get a sum without addends right?