KCC's Quizzes : AQQ234 - Sum of even integers

The general formula for sum of even numbers is S= n(n+1)

with n = 5298   the sum is 2,80,74,102

n = 1; f(n) = 2 = 1 + 1^2

n = 2; f(n) = 6 = 2 + 2^2

n = 3' f(n) = 12 = 3 + 3^2

n = 4; f(n) = 20 = 4 + 4^2

...

So general formula is f(n) = n + n^2

Sum with n = 5298 is therefore 28,074,102

1. S(n)=(n/2)*(n/2+1), n=0, 2, 4, 6, 8, ....

2. S(n=5298)= 7019850

Oh wait, maybe I haven't quite got this right.  My n is the number in the sequence.  If the n in the question refers to the actual even integer, then I would have to divide this by 2 to get the position in the sequence.  In that case, the answer would be 7,019,850, as 5298 is the 2649th number in the sequence of even integers >0.  In this case I concur with Gaetano.

  f(n=2m) = S = 2+4+6+ ... +2m    // by definition of f and change of variable, n= 2m
                = 2*(1 + 2 + 3 + ... + m)   // common factor of 2
               = 2(m*(m+1)/2) // by sum of the m first integers
                = m(m+1)  // simplification
               = (n/2)((n/2)+1) // by our definition of m

f(5298) = f( 2* 2649) = 2649*2650 = 7 019 850

<NOPE, I BELIEVE THIS IS INCORRECT.  I need to go back and spend more time on this.>

I'm noticing that the total given any n, call it T(n), is just n(n+1) as per below...

So, the formula is T(n) = n * (n+1)

For n = 5298, T(n) = 5298 * 5299 = 28,074,102

Not very scientific but I think it's correct.

0.25n^2+0.5n

n=5298 ==> sum is 7019850

2+4+6+8+10.....n then 2(1+2+3....n/2)...so sum of n/2 terms will be 2. (n/2(n/2 + 1)) /2 so after calculating we will get (n/2(n/2 + 1)) and so.... n. (n+2) /4......and if n is 5298 then sum will be 7,019,850.

Thanks Rajesh for your prompt reply!

Well, not sure if n(n+1) can be rigtht!

Look, take the example of n=4. You should have S= 2 + 4 = 6 but you actually get 20 which is wrong...