# Summation of $$k ^ n$$ Lead the Formula

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*}

## Square Summation

( k^3 – (k-1)^3 = – 3 k^2 + 3k – 1 ) , then

begin{eqnarray*} n^3 & = & – 3 sum_{k=1}^n k^2 + 3 sum_{k=1}^n k – n & = & – 3 sum_{k=1}^n k^2 + frac{3}{2} n(n+1) – n 3 sum_{k=1}^n k ^ 2 & = & n^3 + frac{3}{2} n (n+1) – n & = & frac{1}{2} n left{ 2n^2 + 3(n+1) – 2 right} & = & frac{1}{2} n left{ 2 (n + 1) (n – 1) + 3(n+1)right} & = & frac{1}{2} n (n+1) left{ 2 (n – 1) + 3 right} & = & frac{1}{2} n (n+1) (2 n + 1) sum_{k=1}^n k ^ 2 & = & frac{1}{6} n (n+1) (2 n + 1) . end{eqnarray*}

In another way.

As we see in ( n = 1 ) , sum of ( m^2 – (m-1)^2 ) in ( m = 1 ) to ( m = k ) is ( k^2 ) .

[ k^2 = sum _{k=1}^k (2m-1) ]

Sum up above equation of ( k = 1 ) to ( k = n ) .

[ sum _{k = 1} ^n k^2 = sum _{k=1}^n sum _{m=1}^k (2m-1) ]

Check out how many times ( 2m – 1 ) appears. When ( m = 1 ) , ( 2 cdot 1 – 1 ) , it appears ( n ) times for left hand calculation. For right hand Σ(sigma) calculation, it appears in ( k = 1, cdots , n ) .

( m = 2 ) のときの値 ( 2 cdot 2 – 1 ) は 右辺の総和を計算する際に、 ( n – 1 ) 回 出現します。 右辺の外側の Σ(シグマ) を計算するときの ( k = 2, cdots , n ) で出現します。

So, ( 2 m – 1 ) appears ( n + 1 – m ) times for right hand calculation. Therefore we can arrange the above formula as follows.

begin{eqnarray*} sum _{k=1}^n k^2 & = & sum _{m=1}^n (n+1-m)(2m-1) & = & sum _{k=1}^n (n+1-k)(2k-1) end{eqnarray*}

Like I wrote ( sum _{k=1}^n k ) , ( sum _{k=1}^n k^2 = sum _{k=1}^n (n+1-k)^2 ) . Sum up both hands to the preceding equation.

begin{eqnarray*} 3 sum _{k=1}^n k^2 & = & sum _{k=1}^n (n+1-k)(2k-1) + 2 sum _{k=1}^n (n+1-k)^2 & = & sum _{k=1}^n left{ (n+1-k)(2k-1) + 2(n+1-k)^2 right} & = & sum _{k=1}^n left[ (n+1-k) left{ (2k-1) + 2(n+1-k) right} right] & = & sum _{k=1}^n left{ (n+1-k) (2n + 1) right} & = & (2n+1) sum _{k=1}^n (n+1-k) & = & (2n+1) sum _{k=1}^n k & = & (2n+1) left{ frac{1}{2} n (n+1) right} & = & frac{1}{2} n(n+1)(2n+1) sum _{k=1}^n k^2 & = & frac{1}{6} n(n+1)(2n+1) end{eqnarray*}

Finally, we leaded the formula. この式展開の方法は、 2乗和について石を敷き詰めるやり方を表しています。 ここでは詳しい説明は省きますが、 興味がありましたら書籍 代数を図形で解く をご覧ください。