Question 427328
find the sum of the multiples of 3 from 3 to 99 inclusive
<pre><font face = "DFKai-SB" color = "indigo" size = 4><b>
Two ways:  with and without formulas:

1. With formulas:

a<sub>n</sub> = a<sub>1</sub> + (n - 1)d
S<sub>n</sub> = (n/2)(a<sub>1</sub> + a<sub>n</sub>)

a<sub>1</sub> = 3
 d = 3
 
a<sub>n</sub> = a<sub>1</sub> + (n - 1)·d
99 =  3 + (n - 1)·3
99 =  3 + 3(n - 1)
96 = 3(n - 1)
32 = n - 1
33 = n

 S<sub>n</sub> =  (n/2)(a<sub>1</sub> + a<sub>n</sub>)
S<sub>33</sub> = (33/2)(3 + 99)
S<sub>33</sub> = (33/2)(102)
S<sub>33</sub> = 1683 

2:  Without formulas:

 S =    3 +  6 +  9 + ... + 97 + 98 + 99
 S = 3( 1 +  2 +  3 + ... + 31 + 32 + 33)
 S = 3(33 + 32 + 31 + ... +  3 +  2 +  1)
2S = 3(34 + 34 + 34 + ... + 34 + 34 + 34)
2S = 3(34·33)
2S = 3366
 S = 1683

Edwin</pre>