Question

Show that the sum of the first 2n natural numbers is n(2n+1)

Not really sure if there’s a typo in my book or not

Answer 1

This sum is the sum of an arithmetic sequence. There is a formula for the sum of an arithmetic sequence which can be looked up or derived by a variety of means.

A nice approach for this sequence is the following. Notice that the sum of first and last number in the sequence is the same as the sum of the second and second last, and also the same as the sum of the third and third last, and so on.

There are n of these pairs. So the desired sum is n x (first number + last number). But the first number is 1 and the last on is 2n. Thus the desired sum is n(1 + 2n).

Hope this helps!!

Mark Brainleast!!!!!!!!!!!

Answer 2

Consider the set {1, 2, 3, 4, ..., m} where m is some natural number, i.e. a positive whole number.

Now let's say we wanted to add everything in that set.

What we can do is write out the terms

1, 2, 3, 4, ..., m

then just underneath, we write the reverse of that set

m, ..., 4, 3, 2, 1

Here's an example of what I mean. I'll use m = 7

`\begin{array}c \cline{1-7}1 & 2 & 3 & 4 & 5 & 6 & 7\\\cline{1-7}7 & 6 & 5 & 4 & 3 & 2 & 1\\\cline{1-7}\end{array}`

For each column, add straight down

The third row will be nothing but 8's

`\begin{array}c \cline{1-7}1 & 2 & 3 & 4 & 5 & 6 & 7\\\cline{1-7}7 & 6 & 5 & 4 & 3 & 2 & 1\\\cline{1-7}8 & 8 & 8 & 8 & 8 & 8 & 8\\\cline{1-7}\end{array}`

We can see there are seven copies of '8' being added here, so the sum of all the numbers in the set {1,2,3,4,5,6,7} is 8*7 = 56. But wait, we double-counted each item when we formed the second row. So we must divide by 2 to fix the double-counting error.

Therefore, 1+2+3+4+5+6+7 = 56/2 = 28 instead.

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That previous example showed that adding from 1 to 7 means we have 7 copies of (1+7). In other words, 7 copies of 8 being added.

We could write that as 7(1+7)

Then divide that by 2 to fix the double-counting error.

We have 7(1+7)/2

Now imagine that we applied all of what was mentioned to the set {1,2,3,4,5,6,7,8}. Instead of 7's, we use 8's this time to get 8(1+8)/2

Then we can extend it out to the set {1,2,3,4,..., m} and we get the formula m(1+m)/2

We have m copies of (1+m) being added; hence the m(1+m). Don't forget to divide by 2 to fix the double-counting error.

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The key takeaway here is that summing 1+2+3+4+...+m gets us m(1+m)/2

In this case, m = 2n because we want the first 2n natural numbers.

Therefore,

m(1+m)/2

2n(1+2n)/2

n(1+2n)

n(2n+1)

represents the sum 1+2+3+...+n+...+2n

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Try out a value like n = 5 to find that 2n = 2*5 = 10, and,

1+2+3+4+5+6+7+8+9+10 = 55

also

n(2n+1) = 5(2*5+1) = 5*(11) = 55

Showing that the formula works for n = 5. I'll let you try other values of n.