Steven’s notebook from 2026 January 12th.
He decided to use a new notation to express the sumation of terms. Below includes the extracted info and the original hand writen notes

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Basic Sums of Powers

Definition

We define a summation operator for a function $f(x)$f(x):$$S(f) = \sum_{x=1}^{n} f(x)$$S(f)=x=1∑n​f(x)

Our goal is to compute sums of the form:$$\sum_{x=1}^{n} x^k$$x=1∑n​xk

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1. Constant Function

$$S_0 = \sum_{x=1}^{n} 1 = n$$S0​=x=1∑n​1=n

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2. Linear Function

$$S_1 = \sum_{x=1}^{n} x$$S1​=x=1∑n​x

The pairing trick

$$\sum_{x=1}^{n} x = \frac{1}{2} \sum_{x=1}^{n} \bigl(x + (n+1-x)\bigr)$$x=1∑n​x=21​x=1∑n​(x+(n+1−x))

Since each pair sums to $n+1$n+1, we obtain:$$S_1 = \frac{n(n+1)}{2}$$S1​=2n(n+1)​

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3. A Key Telescoping Identity

Consider the difference:$$x^k – (x-1)^k$$xk−(x−1)k

Summing from $x = 1$x=1 to $n$n:$$\sum_{x=1}^{n} \bigl(x^k – (x-1)^k\bigr) = n^k$$x=1∑n​(xk−(x−1)k)=nk

This identity is the foundation of the method.

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4. Binomial Expansion

Using the binomial theorem:$$(x-1)^k = \sum_{i=0}^{k} \binom{k}{i} x^i (-1)^{k-i}$$(x−1)k=i=0∑k​(ik​)xi(−1)k−i

Therefore,$$x^k – (x-1)^k = -\sum_{i=0}^{k-1} \binom{k}{i} (-1)^{k-i} x^i$$xk−(x−1)k=−i=0∑k−1​(ik​)(−1)k−ixi

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5. Recurrence Formula for Power Sums

Summing both sides and rearranging gives:$$\sum_{x=1}^{n} x^k = \frac{1}{k+1} n^{k+1} – \sum_{i=0}^{k-1} \frac{1}{k+1} \binom{k+1}{i} \sum_{x=1}^{n} x^i$$x=1∑n​xk=k+11​nk+1−i=0∑k−1​k+11​(ik+1​)x=1∑n​xi

This expresses a power sum in terms of lower-degree sums.

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6. Example: Sum of Squares

Let $k = 2$k=2.$$\sum_{x=1}^{n} x^2 = \frac{1}{3} n^3 – \frac{1}{3} \left( \binom{3}{0} \sum 1 + \binom{3}{1} \sum x \right)$$x=1∑n​x2=31​n3−31​((03​)∑1+(13​)∑x)

Substituting known sums:$$= \frac{1}{3} n^3 – \frac{1}{3} \left( n + 3 \cdot \frac{n(n+1)}{2} \right)$$=31​n3−31​(n+3⋅2n(n+1)​)

Simplifying:$$\sum_{x=1}^{n} x^2 = \frac{1}{3} n^3 + \frac{1}{2} n^2 + \frac{1}{6} n$$x=1∑n​x2=31​n3+21​n2+61​n

Which factors as:$$\sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6}$$x=1∑n​x2=6n(n+1)(2n+1)​

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7. Summary

• Constant sums are linear in $n$n
• Linear sums are quadratic in $n$n
• Power sums satisfy a recursive formula
• All polynomial sums can be derived using telescoping + binomial expansion