Summing Series -2 questions.

• May 28th 2008, 12:38 PM
Nyoxis
Summing Series -2 questions.
Hey, I've got 2 unsolved problems that need help!

1. Find the sum of the multiples of 7 which are less than 10000.
My answer is 714264285 but the book says 7142142?

2. Find the sum of the series n + 2(n-1) + 3(n-2) +...+ n.
A step by step guide would be really appreciated for this one.

• May 28th 2008, 12:44 PM
Moo
Hello,

Quote:

Originally Posted by Nyoxis
Hey, I've got 2 unsolved problems that need help!

1. Find the sum of the multiples of 7 which are less than 10000.
My answer is 714264285 but the book says 7142142?

There are 1428 multiples of 7 between 1 and 10000, because $\frac{10000}{7} \approx 1428.571428571429$.
Therefore, you can write the sum this way :

$S=7 \cdot 1+7 \cdot 2+7 \cdot 3+\dots+7 \cdot 1427+7 \cdot 1428$

$S=7 \left(1+2+\dots+1427+1428\right)$

What's in brackets is the sum of the 1428 first integers.

We know that $1+\dots+n=\frac{n(n+1)}{2}$

So here, $1+2+\dots+1428=\frac{1428 \cdot 1429}{2}=\boxed{1020306}$

Hence $S=7 \cdot 1020306=\boxed{7142142}$

:)

How did you find 714264285 ?
• May 28th 2008, 12:56 PM
Moo
Quote:

Originally Posted by Nyoxis
2. Find the sum of the series n + 2(n-1) + 3(n-2) +...+ n.
A step by step guide would be really appreciated for this one.

Write it this way :

$S=\sum_{k=0}^{n-1} (k+1)(n-k)$

$S=\sum_{k=0}^{n-1} \left[nk+n-k-k^2\right]=\sum_{k=0}^{n-1} (n-1)k+\sum_{k=0}^{n-1} n-\sum_{k=0}^{n-1} k^2$

$S=(n-1) \sum_{k=0}^{n-1} k+n \sum_{k=0}^{n-1} 1-\sum_{k=0}^{n-1} k^2$

We know that $\sum_{k=0}^n k=\sum_{k=1}^n k=\frac{n(n+1)}{2}$. Be careful, here it's n-1, not n.

We also know that $\sum_{k=0}^n k^2=\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$. Be careful here too :)

There may be a simplier way, but I just can't see it..