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Math Help - sum of sequence

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by Mr_Green View Post
    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.
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  3. #3
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    Or you can always use the "old fashioned way." Simple addition :
    1 + 4 + 9 + 16 + ... + 9409 + 9604

    but I wouldn't recommend it!

    -Dan
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  4. #4
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    Hello, Mr_Green!

    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}<br />
\begin{array}{cccc} = \\ = \\ = \\ =\end{array}<br />
\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 \\<br />
n = 2:\;a\cdot2^3 + b\cdot2^2 + c\cdot2 + d \:=\:5 \\<br />
n = 3:\;a\cdot3^3 + b\cdot3^2 + c\cdot3 + d \:=\:14 \\<br />
n = 4:\;a\cdot4^3 + b\cdot4^2 + c\cdot4 + d \:=\:30 \end{array}<br />
\begin{array}{cccc}\Rightarrow \\ \Rightarrow \\ \Rightarrow \\ \Rightarrow\end{array}<br />
\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}<br />
\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}}

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