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No, if you think that you have it, you can always find one that is larger. Take, for example you find an even composite number (all even numbers, except 2 - which is the only even prime, are composite). Call this number N. Now add 2 to N for N+2, which is also even, so it is also a composite. What about odd composites: if you find an odd composite(M), then add 1 to it: odd+odd=even, so M+1 is even, which is also composite. You can keep going forever.
its 5035the summarian notation tells you that(sum of all #from 0 to a number'N'(sum of all #from 0 to N) = (n)+(n-1)+(n-2)+(n-3)+...+(2) +(1) or(sum of all #from 0 to N) = (1)+ (2) + (3) + (4)+...+(n-1)+(n)the two different sums are aligned by columns. now add the two colunms accordingly and you'll get2x(sum of all #from 0 to N)=(n+1)+(n+1)+...+(n+1) (n+1)is added n timesso2x(sum of all #from 0 to N) =n(n+1)(sum of all #from 0 to N) =n(n+1)/2so (sum of 5 to 100) = (sum of 0 to 100)- (sum of 0 to 5)=100(101)/2 - 5(6)/2 = 5050 - 15 = 5035
For any number n, n + 0 = 0 + n = n All you need to do is to substitute any number of your choice, for n.
N = 3. That really is all there is to it.
Since n! is the product of all the numbers from 1 through n and (n+1)! is everything in n! multiplied by n+1, the quotient is n+1 ■