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Find the value of the sum.

$$\sum_{i=1}^n i(i+1)(i+2)$$

Does this mean that the answer is

$$1(1+1)(1+2) + \cdots + n(n+1)(n+2)$$?

Is there no value to the answer?

Thomas Andrews
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Daniel Valentine
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    Yes, it looks like you're understanding the summation right. No, the sum has no value before you insert a concrete value for $n$. But it happens that the value you then get can also be written as a certain fourth-degree polynomial in the variable $n$ (and no summation); the exercise is to find that polynomial. – hmakholm left over Monica Nov 21 '13 at 23:59
  • What do you mean by the last part of your sentence? – Daniel Valentine Nov 22 '13 at 00:02
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    x @Daniel: The expected answer is something like $42n^4+17n^3+112n^2+5n+9$, only with different coefficients instead of 42, 17, 112, 5, 9. Your task is to find the right coefficients such that the value of the polynomial is also the value of the summation for every $n$. – hmakholm left over Monica Nov 22 '13 at 00:04
  • Is there a way to do this? Sorry I haven't been taught this before. Maybe a video explanation? – Daniel Valentine Nov 22 '13 at 00:08
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    There are several possible methods. One of them, since I have told you you're looking for a degree-4 polynomial, is to compute the summation by hand for $n=1,2,3,4,5$, and then find the (unique) degree-4 polynomial that has the right value at those five points by solving 5 equations in 5 unknowns. Afterwards you can prove by induction that this polynomial indeed gives the right value for all $n$. – hmakholm left over Monica Nov 22 '13 at 00:11

3 Answers3

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Hint: $$4i(i+1)(i+2)= F(i)-F(i-1),$$ where $$F(k)=k(k+1)(k+2)(k+3).$$ This can be verified by writing down $F(i)-F(i-1)$, and taking out common factors. Now add up, and observe the mass cancellations.

The idea generalizes.

André Nicolas
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  • What does writing down F(i) - F(i-1) do? – Daniel Valentine Nov 22 '13 at 00:23
  • When you simplify it, you get $i(i+1)(i+2)$. So $\sum_1^n 4i(i+1)(i+2)=\sum_{i=1}^n F(i)-F(i-1)$. This is $F(1)-F(0)+F(2)-F(1)+\cdots +F(n)-F(n-1)$. There is mass cancellation (telescoping) and we get $F(n)-F(0)$. This gives us a simple closed form formula for our sum. (Well, we have to divide by $4$.) There is an alternate similar argument that uses the binomial coefficient $\binom{n+3}{4}$, and combinatorial ideas. – André Nicolas Nov 22 '13 at 01:05
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Hint:

$$\sum_{i=1}^n 1 = n$$

$$\sum_{i=1}^n i = \frac12n(n+1)$$

$$\sum_{i=1}^n i(i+1) = \frac13n(n+1)(n+2).$$

David H
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You can also solve this problem without any knowledge of summation identities if you're willing to differentiate the function $f$ three times, where

$$f(x)= \frac{x^{n+3}-x^3}{x-1}=\sum_{i=1}^n x^{i+2}$$ This is because

$$f'''(x)=\sum_{1=1}^n [(i+2)(i+1)i]\,x^{i-1}$$ so $f'''(1)$ will give you the final answer.

Doc
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