Sequences & Series: Express, in terms of k, the sum from

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Josephine

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Let k be a positive integer. Find, in terms of k, an expression for S1, the sum of the integers from 2k to 4k inclusive.

Answer: S1 = 3k(2k + 1)

My working:

. . .S1 = (1/2) (2k + 4k)
. . . . . .= k + 2k
. . . . . .= 3k

How can I get 3k(2k + 1)?

Thank you!

Josephine
 
Find the sum for 4k and subtract the sum for 2k-1.

\(\displaystyle \L\\\sum{2k-1}=\frac{(2k-1)(2k)}{2}\)

\(\displaystyle \L\\\sum{4k}=\frac{4k(4k+1)}{2}\)

\(\displaystyle \L\\\frac{4k(4k+1)}{2}-\frac{(2k-1)2k}{2}=3k(2k+1)\)
 
Re: Sequences & Series: Express, in terms of k, the sum

Hello, Josephine!

Let \(\displaystyle k\) be a positive integer.
Find, in terms of \(\displaystyle k\), an expression for \(\displaystyle S\), the sum of the integers from \(\displaystyle 2k\) to \(\displaystyle 4k\) inclusive.

Answer: \(\displaystyle \,S \:= \:3k(2k\,+\,1)\)

My working:
. . .\(\displaystyle S \:= \:\frac{1}{2}(2k + 4k)\) . . . no
Click to expand...

You left out \(\displaystyle n\), the number of terms.

From \(\displaystyle 2k\) to \(\displaystyle 4k\) (inclusive), there are \(\displaystyle 2k\,+\,1\) terms.

Therefore: \(\displaystyle \L\,S\;=\;\frac{2k\,+\,1}{2}\left(2k\,+\,4k\right)\:=\:\frac{2k\,+\,1}{2}(6k)\:=\:3k(2k\,+\,1)\)

 
Oh, I see I forgot the put in n. But, it still confuses me as to why n is 2k+1?

Josephine
 
See for yourself. Let k equal some integer.

Let k=3. Now, we're adding from 2k to 4k. From 6 to 12

6+7+8+9+10+11+12

Count them, 7 terms.

2k+1--->2(3)+1=7

Because of the inclusivity(is that a word?).

4k-2k+1=2k+1

If we were adding from 2k to, say, 5k.

We'd have 3k+1 terms

Let k = 3

6+7+8+9+10+11+12+13+14+15

10 terms

3(3)+1=10
 
Ok, I see it now.
There's also another way.

2k,3k,4k

Un = a + (n-1)d
4k = 2k +(n-1)(1)
4k = 2k + n -1
2k = n-1
n = 2k + 1

Anyway, thanks galactus!
 
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