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series of 1 + 1/2 + 1/3 ...etc
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I need some help: how can I do this:
Write a program to find out how many terms of the series:
1 + 1/2 + 1/3 + 1/4 +.... are needed before the total first exceeds 5.
Write a program to find out how many terms of the series:
1 + 1/2 + 1/3 + 1/4 +.... are needed before the total first exceeds 5.
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The first step to figuring out how to write the program is figuring out how to write the algorithm. How would you solve this problem with pencil and paper? Where would you start? What steps would you take?
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Join Date: Apr 2004
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Use floating point, otherwise the answer is 1.
1 + 1/2.0 + 1/3.0 + 1/4.0
u need an array
float x[20];
// to enter the codes now.use a for loop
int i,y=1;
for (i=0;i<20,i++)
x[i]=1/y+1/y+1;
// to calculate total
for(i=0;i<20;i++)
total=x[i]+x[i+1];
if (total>5) num=x[i]// declare num first somewhere in the program
// how many terms needed
of course this is not the absolute code but i guess you are intellighent enough to go ahead
before writing any program do it on paper first
get the logic first then use your computer
:lol:
float x[20];
// to enter the codes now.use a for loop
int i,y=1;
for (i=0;i<20,i++)
x[i]=1/y+1/y+1;
// to calculate total
for(i=0;i<20;i++)
total=x[i]+x[i+1];
if (total>5) num=x[i]// declare num first somewhere in the program
// how many terms needed
of course this is not the absolute code but i guess you are intellighent enough to go ahead
before writing any program do it on paper first
get the logic first then use your computer
:lol:
:lol: I am not one of those who wait for things to happen, :p but one of those who make things happen ;)
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Join Date: Apr 2004
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>u need an array
No, you don't. In fact, it would be better not to.
>int i,y=1;
>for (i=0;i<20,i++)
>x[i]=1/y+1/y+1;
Reread my previous post.
No, you don't. In fact, it would be better not to.
>int i,y=1;
>for (i=0;i<20,i++)
>x[i]=1/y+1/y+1;
Reread my previous post.
well dave please clarify the reason. coz i am not an expert in c++. thanks in advance
:lol: I am not one of those who wait for things to happen, :p but one of those who make things happen ;)
all you need is a loop, a float to hold the result, and a loopcounter 
I wrote the code in about 3 minutes, including the time to start my editor and the time to compile it.
I wrote the code in about 3 minutes, including the time to start my editor and the time to compile it.
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Originally Posted by anastacia
well dave please clarify the reason. coz i am not an expert in c++. thanks in advance
Or the integer math? In integer math, 1 / 2 = 0. But floating point math, 1.0 / 2 = 0.5. If you take the integer result 0 and (implicitly) convert to floating point, you will have 0.0. You can see how this would seriously affect the calculation.
yeah sort of understanding a little bit. jwenting said something about a counter can you please clarify that ... point.
thnx in advance
thnx in advance
:lol: I am not one of those who wait for things to happen, :p but one of those who make things happen ;)
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Join Date: Apr 2004
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Originally Posted by anastacia
yeah sort of understanding a little bit. jwenting said something about a counter can you please clarify that ... point.
thnx in advance
This is hard not to completely give away since it's only about 4 lines of code -- why not post an attempt?
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