Prove 1 + 4 + 9 + ... + n2 = n (n + 1) (2n + 1) / 6 for all positive integers n.
Another way to write "for every positive integer n" is for every positive integer n. This works because Z is the set of integers, so Z+ is the set of positive integers. The upside down A is the symbol for "for all" or "for every" or "for each" and the symbol that looks like a weird e is the "element of" symbol. So technically, the statement is saying "for every n that is an element of the positive integers", but it's easier to say "for every positive integer n".
On the left hand side, Sk + ak+1 means the "sum of the first k terms" plus "the k+1 term", which gives us the sum of the first k+1 terms, Sk+1.
This often gives students difficulties, so lets think about it this way. Assume k=10. Then Sk would be S10, the sum of the first 10 terms and ak+1 would be a11, the 11th term in the sequence. S10 + a11 would be the sum of the 10 terms plus the 11th term which would be the sum of the first 11 terms.
On the right hand side, replace ak+1 by ( k+1)2, which is what you found it was before beginning the problem.
Sk+1 = k ( k + 1 ) ( 2k + 1 ) / 6 + ( k + 1 )2
Now, try to turn your right hand side into goal of ( k + 1 ) ( k + 2 ) ( 2k +3 ) / 6.
You need to get a common denominator, so multiply the last term by 6/6.
Sk+1 = k ( k + 1 ) ( 2k + 1 ) / 6 + 6 ( k + 1 )2 / 6
Now simplify. It is almost always easier to factor rather than expand when simplifying. This is especially aided by the fact that your goal is in factored form. You can use that to help you factor. You know that you want a (k+1) (k+2) (2k+3) in the final form. We see right now that there is a (k+1) that is common to both of those, so let's begin by factoring it out.
Sk+1 = ( k + 1) [ k ( 2k + 1 ) + 6 ( k + 1 ) ] / 6
What's left inside the brackets [ ] doesn't factor, so we expand and combine like terms.
Sk+1 = ( k + 1) ( 2k2 + k + 6k + 6 ) / 6
Sk+1 = ( k + 1) ( 2k2 + 7k + 6 ) / 6
Now, try to factor 2k2 + 7k + 6, keeping in mind that you need a (k+2) and (2k+3) in the goal that you don't have yet.
Sk+1 = ( k + 1) ( k + 2 ) ( 2k + 3 ) / 6
Hey! That's what our goal was. That's what we were trying to show. That means we did it!
Conclusion
The sum of a constant times a function is the constant times the sum of a function.
The sum of a sum is the sum of the sums
∑(x+y) = ∑x + ∑y
The summation symbol can distribute over addition.
The sum of a difference is the difference of the sums
∑(x-y) = ∑x - ∑y
The summation symbol can distribute over subtraction.
Sum of the Powers of the Integers
The closed form for a summation is a formula that allows you to find the sum simply by knowing the number of terms.
Finding Closed Form
Find the sum of : 1 + 8 + 22 + 42 + ... + (3n2-n-2)
The general term is an = 3n2-n-2, so what we're trying to find is ∑(3k2-k-2), where the ∑ is really the sum from k=1 to n, I'm just not writing those here to make it more accessible.
Replace each summation by the closed form given above. The closed form is a formula for a sum that doesn't include the summation sign, only n.
3[n(n+1)(2n+1)/6] - [n(n+1)/2] - 2[n]
Now get a common denominator, in this case, 2.
n(n+1)(2n+1)/2 - n(n+1)/2-4n/2
Remember that the word factor begins with the letter F and anytime you have a choice of doing something in mathematics that starts with the letter F, that's probably where you should start. So, do not expand, factor instead.
The common factor is n so we'll factor that out of each term. The whole expression is over 2.
n [ (n+1)(2n+1) - (n+1) - 4 ] / 2
Now expand inside the brackets [ ].
n [ 2n2 + 3n + 1 - n - 1 - 4 ] / 2
Simplify like terms.
n ( 2n2 + 2n - 4 ) / 2
Notice the common factor of 2 inside the parentheses, let's factor that out.
2 n ( n2 + n - 2 ) / 2
The 2 in the numerator and the 2 in the denominator divide out and we can factor the rest to get the closed form for the sum.
n ( n + 2 ) ( n - 1)
Isn't that beautiful? At this point, we could write a mathematical induction problem similar to those in the book for this problem. It would read ...
Prove:
sum(3k^2-k-2,k,1,n) = n(n+2)(n-1) for every positive integer n
Answers & Comments
Answer:
Prove 1 + 4 + 9 + ... + n2 = n (n + 1) (2n + 1) / 6 for all positive integers n.
Another way to write "for every positive integer n" is for every positive integer n. This works because Z is the set of integers, so Z+ is the set of positive integers. The upside down A is the symbol for "for all" or "for every" or "for each" and the symbol that looks like a weird e is the "element of" symbol. So technically, the statement is saying "for every n that is an element of the positive integers", but it's easier to say "for every positive integer n".
On the left hand side, Sk + ak+1 means the "sum of the first k terms" plus "the k+1 term", which gives us the sum of the first k+1 terms, Sk+1.
This often gives students difficulties, so lets think about it this way. Assume k=10. Then Sk would be S10, the sum of the first 10 terms and ak+1 would be a11, the 11th term in the sequence. S10 + a11 would be the sum of the 10 terms plus the 11th term which would be the sum of the first 11 terms.
On the right hand side, replace ak+1 by ( k+1)2, which is what you found it was before beginning the problem.
Sk+1 = k ( k + 1 ) ( 2k + 1 ) / 6 + ( k + 1 )2
Now, try to turn your right hand side into goal of ( k + 1 ) ( k + 2 ) ( 2k +3 ) / 6.
You need to get a common denominator, so multiply the last term by 6/6.
Sk+1 = k ( k + 1 ) ( 2k + 1 ) / 6 + 6 ( k + 1 )2 / 6
Now simplify. It is almost always easier to factor rather than expand when simplifying. This is especially aided by the fact that your goal is in factored form. You can use that to help you factor. You know that you want a (k+1) (k+2) (2k+3) in the final form. We see right now that there is a (k+1) that is common to both of those, so let's begin by factoring it out.
Sk+1 = ( k + 1) [ k ( 2k + 1 ) + 6 ( k + 1 ) ] / 6
What's left inside the brackets [ ] doesn't factor, so we expand and combine like terms.
Sk+1 = ( k + 1) ( 2k2 + k + 6k + 6 ) / 6
Sk+1 = ( k + 1) ( 2k2 + 7k + 6 ) / 6
Now, try to factor 2k2 + 7k + 6, keeping in mind that you need a (k+2) and (2k+3) in the goal that you don't have yet.
Sk+1 = ( k + 1) ( k + 2 ) ( 2k + 3 ) / 6
Hey! That's what our goal was. That's what we were trying to show. That means we did it!
Conclusion
The sum of a constant times a function is the constant times the sum of a function.
The sum of a sum is the sum of the sums
∑(x+y) = ∑x + ∑y
The summation symbol can distribute over addition.
The sum of a difference is the difference of the sums
∑(x-y) = ∑x - ∑y
The summation symbol can distribute over subtraction.
Sum of the Powers of the Integers
The closed form for a summation is a formula that allows you to find the sum simply by knowing the number of terms.
Finding Closed Form
Find the sum of : 1 + 8 + 22 + 42 + ... + (3n2-n-2)
The general term is an = 3n2-n-2, so what we're trying to find is ∑(3k2-k-2), where the ∑ is really the sum from k=1 to n, I'm just not writing those here to make it more accessible.
Replace each summation by the closed form given above. The closed form is a formula for a sum that doesn't include the summation sign, only n.
3[n(n+1)(2n+1)/6] - [n(n+1)/2] - 2[n]
Now get a common denominator, in this case, 2.
n(n+1)(2n+1)/2 - n(n+1)/2-4n/2
Remember that the word factor begins with the letter F and anytime you have a choice of doing something in mathematics that starts with the letter F, that's probably where you should start. So, do not expand, factor instead.
The common factor is n so we'll factor that out of each term. The whole expression is over 2.
n [ (n+1)(2n+1) - (n+1) - 4 ] / 2
Now expand inside the brackets [ ].
n [ 2n2 + 3n + 1 - n - 1 - 4 ] / 2
Simplify like terms.
n ( 2n2 + 2n - 4 ) / 2
Notice the common factor of 2 inside the parentheses, let's factor that out.
2 n ( n2 + n - 2 ) / 2
The 2 in the numerator and the 2 in the denominator divide out and we can factor the rest to get the closed form for the sum.
n ( n + 2 ) ( n - 1)
Isn't that beautiful? At this point, we could write a mathematical induction problem similar to those in the book for this problem. It would read ...
Prove:
sum(3k^2-k-2,k,1,n) = n(n+2)(n-1) for every positive integer n
: