Question

Find the sum of the following: First 20 positive integers ending in 3

Answer 1

3, 13, etc...
Rule: 3+10(n-1) = 10n-7, n ranges from 1 to 20.
Sum of that = (10*20-7 +3)*10/2 = 1960/2 = 980

Answer 2

:D

Answer 3

but is the first 20 Positive integers . . .

Answer 4

and it says the answer is 1960

Answer 5

i will put my money on daniel
heh but not sure how he got that formula sorry

Answer 6

haha okay thanks :)

Answer 7

u got the answer?

Answer 8

I divided by 2 twice. It should have been 196*20/2, not 196*10/2 (I second-guessed myself). Your book is right :D

Answer 9

10(6 + 19(10))

Answer 10

The formula is an adaptation old Gaussian equation n(n+1)/2.
He did it by saying that:
1 + 2 + 3 + 4 + ... + n-1 + n
+ n + n-1 + ... + 4 + 3 + 2 + 1
= (n+1)n/2

What I did:
3 + 13 + 23 + ... + 183 + 193
+193 + 183 + ... + 23 + 13 + 3
=196*(20)/2 = 1960

Answer 11

sum of an arithmetic progression
(n/2) [2a + (n-1)d]

Answer 12

^ Yup. That's the generalized version. Cleaned up a little bit:\[n/2 (2a + (n-1)d)\]

Answer 13

Sorry about the first answer UkaLailey. I'm going to start a thread that asks if I should go into retirement... I've missed 2 today...

Answer 14

hahahaha...lol

Answer 15

you deserved a medal for that one :P