Find the Sum

magentarita · #1 $Old$ 11-19-2008, 07:52 PM

What is the sum of all multiples of 6 that are strictly between 3,128 and 51,804?

osana412 · #2 $Old$ 11-19-2008, 10:42 PM

The answer 65,812,656 it is true?

look bellow, it is correct?

magentarita · #3 $Old$ 11-20-2008, 07:06 AM

Quote:

Originally Posted by osana412 $View Post$

The answer 65,812,656 it is true?

look bellow, it is correct?

I don't know if the answer is ok.

I'll get back to you.

Soroban · #4 $Old$ 11-20-2008, 11:55 AM

Hello, magentarita!

Quote:

What is the sum of all multiples of 6 that are strictly between 3,128 and 51,804?

We have: . $S \;=\;3,\!132 + 3,\!138 + 3,\!144 + \hdots + 51,\!798$

. . . . . . $= \;6\cdot522 + 6\cdot523 + 6\cdot524 + \hdots + 6\cdot8633$

. . . . . . $= \;6\underbrace{\bigg[522 + 523 + 524 + \hdots + 8633\bigg]}_{\text{arithmetic series}}$

The series has: first term $a = 522$ , common difference $d = 1$ , and $n = 8112$ terms.
Its sum is: . $\tfrac{8112}{2}[2(522) + 8111(1)] \:=\:37,\!132,\!680$

Therefore: . $S \;=\;6 \times 37,\!132,\!680 \;=\;\boxed{222,\!796,\!080}$

magentarita · #5 $Old$ 11-20-2008, 04:23 PM

Quote:

Originally Posted by Soroban $View Post$

Hello, magentarita!

We have: . $S \;=\;3,\!132 + 3,\!138 + 3,\!144 + \hdots + 51,\!798$

. . . . . . $= \;6\cdot522 + 6\cdot523 + 6\cdot524 + \hdots + 6\cdot8633$

. . . . . . $= \;6\underbrace{\bigg[522 + 523 + 524 + \hdots + 8633\bigg]}_{\text{arithmetic series}}$

The series has: first term $a = 522$ , common difference $d = 1$ , and $n = 8112$ terms.
Its sum is: . $\tfrac{8112}{2}[2(522) + 8111(1)] \:=\:37,\!132,\!680$

Therefore: . $S \;=\;6 \times 37,\!132,\!680 \;=\;\boxed{222,\!796,\!080}$

You have no idea how much math is sinking into my head thanks to your replies.