The second one reminds me of a joke that someone told me once:
What is the difference between a psychotic, a neurotic, and a mathematician?
A psychotic believes that 2+2=5, and neurotic believes that 2+2=4 but it kills him, and a mathematician just changes the base. :-) Sorry. Don't know if that was what you were looking for, but a little more humor nonetheless.
lol... actually the answer was:
If the scientist ignored significant figures and the 2's were actually 2.4's then 2.4 + 2.4 would = 5!
There are two significant figures in 2.4
By the way, the sum of the first n integers is .5(n)(n+1), not (n+2)^2, so the sum of the first 104 natural numbers is 5 460.
When n = 1, answer is 9 due to 1+2 squared.
When n = 2, answer is 16 due to 4 squared. Meaning the sum of the first two digits is 25....
I think we're trying to calculate different values. What do you mean by "the first n digits"?
By the way, the sum of the first n integers is .5(n)(n+1), not (n+2)^2, so the sum of the first 104 natural numbers is 5 460.
This is correct, it was just unclear how KJ worded it, that's all.
Part 2 has already been sufficiently answered,
for part 1)
Sum{from n=1 to n=104} ((n+2)^2) = Sum{from n=1 to n=104} (n^2 + 4*n + 4)
Since this is a finite sum, we may split the sum up across the terms without any problems (really we are using the associative property to do this.) So, we have:
Sum{from n=1 to n=104} (4) + 4 * Sum{from n=1 to n=104} (n) + Sum{from n=1 to n=104} (n^2)
Where the second term follows by factoring the 4 out of the sum.
The first term is just 104*4=416
The second term can be understood as the area of a triangle:
*
**
***
****
.
.
.
**************************** <----------------- Bottom row has n astrikes. To account for the fact that we include all of the stars on the hypotenuse, we use the formula:
4*(1/2)*104*105 = 21840 (Where the factor of 4 comes from our original sum- not the area of the rectangle)
The last term can be computed as follows by using a technique called Proof by Mathematical Induction:
CLAIM: For Q>0, Sum{from n=1 to n=Q} (n^2) = Q*(Q+1)*(2Q+1)/6
Proof:
Q=1 case: 1*2*3/6=1
Q=2 case: 2*3*5/6=5
So we suspect that it works for all natural numbers N, now we will assume it works for some arbitrary natural number K and show that this forces the formula to work for K+1.
Assume that Sum{from n=1 to n=K} (n^2) = K*(K+1)*(2K+1)/6.
Then Sum{from n=1 to n=K+1} (n^2) = 1 + 2^2 + 3^2 + ... + K^2 + (K+1)^2
We have already assumed that 1 + 2^2 + 3^2 + ... + K^2 = K*(K+1)*(2K+1)/6, so the sum immediately above this one equals:
K*(K+1)*(2K+1)/6 + (K+1)^2 = K*(K+1)*(2K+1)/6 + (6/6)*(K+1)^2 =
(K*(K+1)*(2K+1)+6*(K+1)^2)/6 = ((K+1)(K*(2K+1)+6*(K+1)))/6 =
... after foiling out the terms in second factor, (K*(2K+1)+6*(K+1)) and then refactoring, we get:
((K+1)(K+2)(2*(K+1)+1))/6 which is the form we wanted.
Since we have shown that this property is true for N=1, and that if the property is true for any arbitrary N implies that it must be true for N+1 (Hence, its truth for 1 implies the truth for 2 implies the truth for 3, etc...) So this property must be true for any natural number N.
So, letting N=104, we have that Sum{from n=1 to n=104} (n^2) = (104)*(104+1)*(2*(104+1)+1)/6 = 380380
So our original sum is:
Sum{from n=1 to n=104} ((n+2)^2) = 416+21840+380380 = 402636
Yup, it's definitely mathematical induction. I didn't get the part about the asterisks and stars on a hypotenuse. Could you please clarify for my ignorance?
You're right, that was a lame analogy. I will either draw a picture and scan it to clairify what I mean or think of a better way of arguing that point.
Ok, I am a dork, this is a much clearer way of thinking about that n*(N+1)/2 identity. Consider the case K=4
%@@@@
%%@@@
%%%@@
%%%%@
Notice that if we add up the number of percent signs, we get 1+2+3+4 = 10 This is the same as computing the area of the rectangle made up of both percent signs and @ symbols and then divide by 2 (since the area of the portion of the rectangle made up of % is equal to the area of the portion of the rectangle made up @ symbols). The area of this rectangle (regarding each symbol as having unit length) is just 4*5 = 20 so one half this is 10 which is the area of either the % triangle (the one we care about) or the @ triangle (the one we dont care about).
I hope this clairifies this... let me know if I can further clairify.
Ok, I am a dork, this is a much clearer way of thinking about that n*(N+1)/2 identity. Consider the case K=4
%@@@@
%%@@@
%%%@@
%%%%@
Notice that if we add up the number of percent signs, we get 1+2+3+4 = 10 This is the same as computing the area of the rectangle made up of both percent signs and @ symbols and then divide by 2 (since the area of the portion of the rectangle made up of % is equal to the area of the portion of the rectangle made up @ symbols). The area of this rectangle (regarding each symbol as having unit length) is just 4*5 = 20 so one half this is 10 which is the area of either the % triangle (the one we care about) or the @ triangle (the one we dont care about).
I hope this clairifies this... let me know if I can further clairify.
Thanks, this helps a lot. Sorry about my deficiencies ;).
1) In this formula, find the Sum of the first 104 digits
(n+2)^2, where n represents the term number. Good luck :P
2) Show how 2+2 could equal 5 if scientists who were calculating data ignored something crucial in calculating data in science....
In number 2) you could just allow for the division by zero.. then I think you would be able to get 2+2=5 ..
Division by 0 always "works" ;)
1) In this formula, find the Sum of the first 104 digits
(n+2)^2, where n represents the term number. Good luck :P
2) Show how 2+2 could equal 5 if scientists who were calculating data ignored something crucial in calculating data in science....