Arithmetic Progression question

Leinsterr

Hi, im in 4th year in secondary school. Im training for the maths olympics atm in ucd and the notes the lecturer gave us were useless. Can someone please try and help me with these questions and give me explanations for the answers. Thanks

Caluclate the sums:

1(i) 1(3)+2(4)+3(5)+...+n(n+2)
(ii) 1(3)(3)+2(4)(6)+3(5)(7)+...+n(n+2)(n+4)

hivizman

I suspect that the questions are assuming that you know the general formulae for the sums of the integers from 1 to n, the squares of the integers from 1 to n, and the cubes of the integers from 1 to n. Given this knowledge, the best approach is to expand the general term and then work out the sums of each element.

For example, in the first question, n(n+2) = n^2 + 2n, so the sum of the terms from 1(3) to n(n+2) is equal to the sum of the squares from 1 to n plus two times the sum of the integers from 1 to n.

n(n+1)(2n+1)/6 + 2*[n(n+1)/2] = n(n+1)(2n+7)/6

The second question can be answered using the same approach, with n(n+2)(n+4) expanded to n^3+6n^2 +8n.

Leinsterr

hivizman wrote: »

I suspect that the questions are assuming that you know the general formulae for the sums of the integers from 1 to n, the squares of the integers from 1 to n, and the cubes of the integers from 1 to n. Given this knowledge, the best approach is to expand the general term and then work out the sums of each element.

For example, in the first question, n(n+2) = n^2 + 2n, so the sum of the terms from 1(3) to n(n+2) is equal to the sum of the squares from 1 to n plus two times the sum of the integers from 1 to n.

n(n+1)(2n+1)/6 + 2*[n(n+1)/2] = n(n+1)(2n+7)/6

The second question can be answered using the same approach, with n(n+2)(n+4) expanded to n^3+6n^2 +8n.

thanks so much for the help. the formulae we were given were:
(a) 1+2+3+...+n=n(n+1)/2
(b) 1+3+5+...+ (2n-1)=n^2
(c) 1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6

I am still a bit lost. Can you try and explain it a bit further if you have the time? Also what does 2* mean? Is that 2 multipled by or 2 squared?

hivizman

Leinsterr wrote: »

thanks so much for the help. the formulae we were given were:
(a) 1+2+3+...+n=n(n+1)/2
(b) 1+3+5+...+ (2n-1)=n^2
(c) 1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6

I am still a bit lost. Can you try and explain it a bit further if you have the time? Also what does 2* mean? Is that 2 multipled by or 2 squared?

I'm using the "*" symbol to indicate multiplication, which is what you would use in Excel. The "^" symbol means "to the power of", also as in Excel. Because we are using a "sans-serif" font here, the "x" that is often used as a multiplication sign is too easily confused with "x" often used to indicate a variable, so I try to avoid this confusion by using "*" for multiplication.

In answering both questions, your aim should be to break down the series you are given into a set of series for which you know the formula.

The first sum is 1(3) + 2(4) + 3(5) + ... + n(n+2)

Can you express each individual term in a form that looks like the last (the "general") term? Well, 1(3) can be expressed as 1(1+2), 2(4) can be expressed as 2(2+2), 3(5) can be expressed as 3(3+2), and so on. Opening up the brackets, we can write these as (1*1+1*2), (2*2+2*2), (3*3+3*2), and so on, up to (n*n+n*2).

But if we collect all the first terms in each of these expressions together, we have 1*1 + 2*2 + 3*3 + ... + n*n, which is just another way of writing:

1^2 +2^2 +3^2 + ... + n^2. You have a formula to calculate this sum.

If we collect the second terms in each of these expressions together, we have 2*1 + 2*2 + 2*3 + ... + 2*n, which is just another way of writing:

2*(1 + 2 + 3 + ... +n). You also have a formula to calculate the sum in the brackets. Putting it all together, you get the expression that I included in the "spoiler" in my previous post.

Have a go yourself at the second question. First of all, see if you can get the expression in my previous post. Then, you may wish to note that the sum of the first n cubes of positive integers is:

1^3 + 2^3 +3^3 + ... + n^3 = [(n^2)((n+1)^2)]/4

(This formula is best remembered as the square of the formula for the sum of the first n positive integers: 1 + 2 + 3 + ... + n = n(n+1)/2.)

Substitute the various formulae into the expression, do some simplification, and you should get

n(n+1)(n+4)(n+5)/4

.