0

I am supposed to find the value of this:

$$ \sum_{i=1}^n i^2 + 3i + 4$$

I'm a bit unsure about how to simplify this.

I know when I plug in numbers, I get:

$$(1+3+4) + (4+6+4) + (9 + 9+ 4) + (16+12+4) = 8 + 14 + 22 + 32$$

I can see the difference between the numbers increases by 2 with the difference between the first and second numbers starting at 6.

But, how do I come up with a general value to this equation using $n$ and $i$?

Here are some sums that I found:

When $n=5, sum = 120$ $n=4, sum = 76$ $n=3, sum = 44$ $n=2, sum = 22$ $n=1, sum = 8$

daw
  • 49,113
  • 2
  • 38
  • 76
Jwan622
  • 5,704
  • tagged as measure-theory?? seriously?? – daw Mar 24 '18 at 13:31
  • @jwan622 Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 24 '18 at 21:27

3 Answers3

1

Hint:

$$\sum_{i=1}^n i^2=\tfrac16(2n^3+3n^2+n)$$

and

$$\sum_{i=1}^n i=\tfrac12(n^2+n)$$

1

HINT

Note that

$$\sum_{i=1}^n i^2 + 3i + 4=\sum_{i=1}^n i^2+3\sum_{i=1}^n i +\sum_{i=1}^n 4$$

For the basic sum refer to How Are the Solutions for Finite Sums of Natural Numbers Derived? and also the trivial $\sum_{i=1}^n 4=4n$.

user
  • 154,566
0

Note that \begin{eqnarray} (i+1)-i &=& 1 \\ (i+1)^2-i^2 &=& 2i+1 \\ (i+1)^3-i^3 &=& 3 i^2+3i+1 \\ \end{eqnarray}

Hence $\sum_{i=1}^n 1 = \sum_{i=1}^n ((i+1)-i) = n $.

Then $\sum_{i=1}^n ( (i+1)^2-i^2 ) = (n+1)^2-1^2 = \sum_{i=1}^n 2i + \sum_{i=1}^n 1 = 2 \sum_{i=1}^n i +n$, and so $\sum_{i=1}^n i = {1 \over 2} ((n+1)^2-1^2 -n)= {1 \over 2} n (n+1)$.

A similar analysis applies to the last and gives $\sum_{i=1}^n i^2 = {1 \over 6} n (n+1) (2n+1)$.

Note the analogy with integrating $1,x,x^2$ with respect to $x$.

copper.hat
  • 172,524