The question is:
$$\sum_{i=1}^n (i^2+3i+4)$$
I get that
$$\sum_{i=1}^n i^2 = \frac{n(n+1)(n+2)}{6}$$ and $$3\sum_{i=1}^n i = \frac{3n(n+1)}{2}$$ so one would get
I'll call this form1: $$\frac{n(n+1)(n+2)}{6} + \frac{3n(n+1)}{2} + 4n$$
However, the textbook that I using says the answer is:
I'll call this form2: $$\frac{n(n^2+6n+17)}{3}$$
So the part I am confused with is the steps in between form1 and form2.
On a last note it been a good year since I've done any calculus so it would appreciated if you would point the relevant concepts so I can review. Thanks.
$\endgroup$ 14 Answers
$\begingroup$There are several mistakes. Below is the clarification.
$$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$ $$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ $$\sum_{i=1}^n 1 = n$$
Hence
\begin{align} \sum_{i=1}^n (i^2+3i+4) & = \sum_{i=1}^n i^2 + 3 \sum_{i=1}^n i + 4\sum_{i=1}^n 1\\ & = \frac{n(n+1)(2n+1)}{6} + 3\frac{n(n+1)}{2} + 4n \\ & = \frac{n}{6}(2n^2 + 3n + 1 + 9n + 9 + 24)\\ & = \frac{n}{6}(2n^2 + 12n + 34)\\ & = \frac{n}{3}(n^2 + 6n + 17) \end{align}
$\endgroup$ $\begingroup$Here's your mistake: $$\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6}$$
$\endgroup$ 1 $\begingroup$The last term $4$ should be $4n$, because you are adding $4$ for $n$ times.
$\endgroup$ 1 $\begingroup$$$\sum_{k\le n} c=cn\neq c$$ In your example, $\displaystyle\sum 4=4n,$ yet you wrote $4$.
EDIT: $$\frac{n(n+1)(n+2)}{6} + \frac{3n(n+1)}{2} + 4n = n(n^2+6n+17)/3$$ $$\frac{n(n+1)(n+2)}{2} + \frac{9n(n+1)}{2} + 12n = n^3+6n^2+17n$$ $$\frac{n^3+3n^2+2n}{2} + \frac{9n^2+9n}{2} = n^3+6n^2+5n$$ $${n^3+3n^2+2n} + {9n^2+9n}= 2n^3+12n^2+10n$$ $${2n} + {9n}= n^3+10n$$ $$1\neq n^2$$ Therefore one of them is incorrect. We see that you messed up on $\sum i^2$, it equals $\frac{n(n+1)(2n+1)}{6}$
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