Math question of the day

[TheFaQQer]

I hate the American way of abbreviating mathematics... even worse when they say it

When you consider the American school curriculum, it actually makes sense. They tend to teach everything on one topic one year, then next year a different topic - so 11th grade may be trigonometry, and 12th is all calculus (or it was where my sister went to school).

[rdglad]

https://en.wikipedia.org/wiki/Arithmetic_progression

S = n/2 [2 x a + (n - 1) x d]

1, 3, 5, 7,..............

Here
n = 75
a = 1
d = 2
S = 75/2 [ 2 x 1 + (75 -1) x 2]
= 75/2 [ 2 x 1 + 74 x 2]
= 75/2 [ 1 + 74 ] x 2
= 75 x 75
= 5625

In case of odd series:

a is always 1 and d is always 2

S = n/2 [ 2 x 1 + (n - 1) x 2]

If you simlify it:
S = n x n
= 75 x 75
= 5625

Mrs Miggins (the class teacher) asked the class to see if they could calculate the sum of the first 35 odd numbers.


The class started to work on the answer and quick as a flash the young sasguru ran to her and said, 'The sum is 1,225.'

Mrs Miggins thought, 'Wow, lucky guess,' and gave young sassy the task of finding the sum of the first 75 odd numbers. Within 10 seconds, sasguru was back with the correct answer.

How did the young, gifted, super intelligent sasguru find the sum so quickly and what is the answer ??

[OwlHoot]

I was messing about in my earlier reply (although that ridiculous method does always work!)

Probably the simplest approach is:

S = 1 + 3 + .. + 2n-1

= (2 - 1) + (4 - 1) + .. + (2n - 1)

= 2(1 + 2 + .. + n) - (1 + 1 + .. + 1)

= n(n + 1) - n

= n^2

edit, although as others mentioned, even simpler is:

2 S = (1 + 2n-1) + (3 + 2n-3) + ...

= 2 n^2

[AtW]

I was expecting to see something like Goldbach conjecture or solving the happy end problem for arbitrary N in this thread but all I found was kindergarden question.

For this reason I am out.