How much is the summation from 1 to 100?
ガウスがこの計算を即座にやってのけたという話はあまりに有名で、整数を順次足す、等差数列を順次足す方法はご存じの方も多いと思います。
よく使われる説明の方法として、石を三角の形に置き、それと点対象な三角の形に石を置くというのがあります。
Then, how about (1^2 + 2^2 + 3^3 + cdots + 100^2) ?
Here, I lead the formula by equations.
Suppose (k) is a number not less than 1. Think about (k^2 – (k-1)^2 ) .
[ k^2 – (k-1)^2 = 2k – 1 ]
Then, sum up the equations of (k=1) to (n) .
begin{eqnarray*}
n^2 – (n – 1)^2 & = & 2n – 1
(n – 1)^2 – (n – 2) ^ 2 & = & 2(n – 1) – 1
& vdots &
2 ^ 2 – 1 ^ 2 & = & 2 cdot 2 – 1
1 ^ 1 – 0 ^ 2 & = & 2 cdot 1 – 1
end{eqnarray*}
Adding each hand, finally, (n^2) is on the left hand, ( 2 sum_{k=1}^n k – n ) is on the right hand.
[ n^2 = 2 sum_{k=1}^n k – n ]
Then, arrange the equation leads to the final formula.
begin{eqnarray*}
n^2 & = & 2 sum_{k=1}^n k – n
2 sum_{k=1}^n k & = & n^2 + n
& = & n(n+1)
sum_{k=1}^n k & = & frac{1}{2} n(n+1)
end{eqnarray*}
In the same way, we can find the formula of squared summation, three times square summation.
Now, let’s try another way.
The summation of 1 to ( n ) and one of ( n ) to 1 is the same. The the following equation will appear.
begin{eqnarray*}
sum _{k = 1} ^n k & = & sum _{k = 1} ^n (n + 1 – k)
& = & sum _{k = 1} ^n (n + 1) – sum _{k = 1} ^n k
2 sum _{k = 1} ^n k & = & sum _{k = 1} ^n (n + 1)
& = & n ( n + 1 )
sum _{k = 1} ^n k & = & frac{1}{2} n (n + 1)
end{eqnarray*}
