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# KCC's Quizzes : AQQ234 - Sum of even integers

The sum S of positive even integers starting from 2 to n (even) is described as :

S = 2 + 4 + 6 + 8 +… n where the last element n is even.

Questions:

1. What is the general formula (function of n) giving a such sum?
2. Give the sum with n = 5298

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[edited by: emassa at 1:29 PM (GMT -4) on 5 May 2023]

## Top Replies

• 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...

 n 1 2 3 4 5 2n 2 4 6 8 10 T(n) 2 6 12 20 30 n(n+1) 2 6 12 20 30

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.