Question

C++,


Write a recursive function “Summation” that takes two integer numbers n and m and it returns the summation of all the numbers between n and m. If n > m the function returns zero.
Examples:
Summation(3, 7) = 3 + 4 + 5 + 6 + 7 = 25
Summation(7, 2) = 0
Summation(4,4) = 4
Summation(-3,1) = -5

Answer 1

what grade is this (I'm in 6th).

Answer 2

im not going to school

Answer 3

are you on flvs?

Answer 4

nope ;p

Answer 5

o.o ?

Answer 6

@ganeshie8

Answer 7

can you help me with the problem plz

Answer 8

ask Pooja 195.

Answer 9

Thanks

Answer 10

@pooja195

Answer 11

I don't know this stuff sorry! ;-;

Answer 12

I see its ok

Answer 13

... sorry.

Answer 14

np

Answer 15

okay so basically

Answer 16

Summation(3, 7)
this 7 has to keep going down by 1 until 3 for each recursion

Answer 17

What type of math is this???

Answer 18

int sum(int n,int m)
{

if ( n>m)
return 0;
else{
return n+sum(n,m-1);
}


}

int main()
{
cout << sum(3,7);
}

Answer 19

closest I could come up with

Answer 20

return m+sum(n,m-1); ?

Answer 21

results got better but still

Answer 22

oh wait it worked

Answer 23

How did you find out lol

Answer 24

If you return n+sum(n,m-1), you just keep adding n for (n-m+1) times.

Answer 25

one more try ;)

Answer 26

sum(a,b)
if b>a
return b+sum(a,b-1)
else
return 0

Answer 27

sum(a,b)
if b>a
return a+sum(a+1,b)
else
return 0

Answer 28

this is okay too if u want to count up in each recursion its more natural

Answer 29

since thats how ur question is stated

Answer 30

\[S_n = a_n + S_{n-1}\]

Answer 31

http://prntscr.com/97l9uh

Answer 32

int sum(int n,int m)
{

if ( n>m)
return 0;
else{
return n+sum(n,m-1);
}


}

sum (n,m)
= n+sum(n,m-1)
= n + (n + sum (n, m-2)) = 2n + sum(n, m-2)
= 2n + (n + sum (n, m-3)) = 3n + sum (n, m-3)
= ...
= kn + (n + sum(n, m-(k+1)) = (k+1)n + sum(n, m-(k+1))
= ...
Until m = n, we have
= (n-1)n + (n + sum(n, m-n) = n*n + sum (n, m-n)
= n*n + (n) + sum(n, m-(n+1))
= (n+1)n + 0
= (n+1)n
So, you add n for n+1 times. Right? Still wrong?

Answer 33

Hey, I think counting up or counting down doesn't really matter. Both ways work. My concern is, how many times you added "n" in the asker's solution.