# sum of sequence

• Feb 3rd 2007, 05:15 PM
Mr_Green
sum of sequence
Find 1^2 + 2^2 + 3^3 + 4^2 +5^2 + ... 97^2 + 98^2.

I only know how to use the sum/seq functions on my graphing calculator to solve this. Could someone show me how to solve this without that cheat.

THANKS
• Feb 3rd 2007, 05:20 PM
ThePerfectHacker
Quote:

Originally Posted by Mr_Green
Find 1^2 + 2^2 + 3^3 + 4^2 +5^2 + ... 97^2 + 98^2.

I only know how to use the sum/seq functions on my graphing calculator to solve this. Could someone show me how to solve this without that cheat.

THANKS

I assume you mean,
$1^2+2^2+3^2+...$
(Not $3^3$).
Thus formula is,
$1^2+2^2+...+n^2=\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$
You can do it from there.
• Feb 4th 2007, 12:31 PM
topsquark
Or you can always use the "old fashioned way." Simple addition :eek: :
1 + 4 + 9 + 16 + ... + 9409 + 9604

but I wouldn't recommend it! :)

-Dan
• Feb 4th 2007, 01:56 PM
Soroban
Hello, Mr_Green!

Quote:

Find $1^2 + 2^2 + 3^3 + 4^2 +5^2 + \cdots + 97^2 + 98^2$

ThePerfectHacker gave you the formula for this problem.
If you're not familiar with it, you could derive it,
. . but it's faster to simply add the ninety-eight squares!

For anyone who is interested, here is one method.

List the first few partial sums.

$\begin{array}{ccccc}1^2 \\ 1^2+2^2 \\ 1^2+2^2+3^2 \\ 1^2 + 2^2+3^2+4^2 \\ \vdots\end{array}
\begin{array}{cccc} = \\ = \\ = \\ =\end{array}
\begin{array}{ccccc} 1 \\ 5 \\ 14 \\ 30 \\ \vdots\end{array}$

Take differences of consecutive pairs of the numbers.
Take differences of the differences, and so on
. . until you reach a row of constant differences.

. . $1\quad5\quad 14\quad 30\quad 55\quad 91 \;\cdots$
. . . $4\quad\,9\quad16\quad25\quad36\;\cdots$
. . . . . $5\quad7\quad\;9\quad\;11\;\cdots$
. . . . . . $2\quad2\quad\;2\;\:\cdots$

The constants appeared in the third differences.
Hence, the generating function is of the third degree (a cubic).

The general form of a cubic is: . $f(n) \:=\:an^3 + bn^2 + cn + d$

We know the first four sums of the sequence; plug them in.
$\begin{array}{cccc}n = 1:\;a\cdot1^3 + b\cdot1^2 + c\cdot1 + d \:=\:1 \\
n = 2:\;a\cdot2^3 + b\cdot2^2 + c\cdot2 + d \:=\:5 \\
n = 3:\;a\cdot3^3 + b\cdot3^2 + c\cdot3 + d \:=\:14 \\
n = 4:\;a\cdot4^3 + b\cdot4^2 + c\cdot4 + d \:=\:30 \end{array}
\begin{array}{cccc}\Rightarrow \\ \Rightarrow \\ \Rightarrow \\ \Rightarrow\end{array}
\begin{array}{cccc}a + b + c + d & =\:1 \\ 8a + 4b + 2c + d & =\:5 \\ 27a + 9b + 3c + d & =\:14 \\ 64a + 16b + 4c + d & =\:30\end{array}
\begin{array}{cccc}(1)\\(2)\\(3)\\(4)\end{array}$

$\begin{array}{ccc}\text{Subtract (1) from (2):} & 7a + 3b + c & =\;4 \\ \text{Subtract (2) from (3):} & 19a + 5b + c & = \;9\\ \text{Subtract (3) from (40:} & 37a + 7b + c & =\:16\end{array}\begin{array}{ccc}(5)\\(6)\\(7)\en d{array}$

$\begin{array}{cc}\text{Subtract (5) from (6):} \\ \text{Subtract (6) from (7):}\end{array}\begin{array}{cc}12a + 2b & =\;5 \\18a + 2b & = \:7 \end{array}\begin{array}{cc}(8)\\(9)\end{array}$

$\text{Subtract (8) from (9): }\;6a\,=\,2\quad\Rightarrow\quad\boxed{a = \frac{1}{3}}$
$\text{Substitute into (8): }\;12\left(\frac{1}{3}\right) + 2b \:=\:5\quad\Rightarrow\quad\boxed{b = \frac{1}{2}}$
$\text{Substitute into (5): }\;7\left(\frac{1}{3}\right) + 3\left(\frac{1}{2}\right) + c\:=\:4\quad\Rightarrow\quad\boxed{c = \frac{1}{6}}$
$\text{Substitute into (1): }\;\frac{1}{3} + \frac{1}{2} + \frac{1}{6} + d \:=\:1\quad\Rightarrow\quad\boxed{d = 0}$

Therefore, the function is: . $f(n) \;=\;\frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n \;=\;\boxed{\frac{n(n+1)(2n+1)}{6}}$