Derivation of the formula for summation of the power of positive integers, the formula for summation

Problem: For k n, accumulate from 1 to m, solve.

Apparently Gaussian has been solved for the first power (sum of natural numbers), and in fact the first power can be reduced to the power of 0 (i.e. 1).

Reduced power: For k n, we want to split it into two parts, which can only be k (n+1)-(k-1) (n+1), after the formula is used by the quadratic term theorem, the power n+1 will cancel out, and the term containing k n is solved as x, which is probably k n=(k (n+1)-(k-1) (n+1)+n(n-1)k (n-1)+....Adding up the two sides of this equation gives k n=m (n+1)-(1-1) (n+1)+....The part represented by the ellipsis is also an additive, but exponential compared to the original.

For powers of any order, the problem can be solved by successively decreasing the power to 0 or 1.

If we add from j to i, then we can first find the sum from 1 to i and then subtract the sum from 1 to j.

k=1. sn=n(n+1)/2

k=2. sn=n(n+1)(2n+1) 6k=3. sn=[n(n+1)]^2/4

The sn expression of k=2 is derived from the expression of k=1, and the sn expression of k=3 is derived from the expression of k=2. That is, we now know the expression at k=1, and we know the relationship between k=m and k=m-1 expression (recursive), so we can know all the expressions of kn.

But so far, I haven't seen the general formula of the SN about k.

This general formula may or may not exist.

If you must know the general formula, just guess with the formulas k=1,2,3 and prove it by mathematical induction.

The sum formula for the power function is: s=n+(n-1)+(n-2)+1, where all the added binomial formulas are determined according to the following binomial formulas, so that the progressive derivation of the summation formula of the natural numbers from 1 to n powers can be smoothly carried out.

The process of derivation: The formula of the binomial theorem can be transformed into a series of equal differences, from the lower power to the higher power, and finally it can be deduced to the original form of Li Shanlan's natural power summation formula.

When n is an odd number, it is determined by 1+2+3+.n and s = n + (n-1) + (n-2) +1 Sum up:

2s=n+[1+(n-1)]+2+(n-2)]+3+(n-3)]+n-1)+(n-n-1)]+n=n+n+n+..n plus or minus all added binomial formulas = (1+n)n minus all added binomial formulas.

When n is an even number, it is determined by 1+2+3+.n and s = n + (n-1) + (n-2) +1 Sum up:

2s=n+[1+(n-1)]+2+(n-2)]+3+(n-3)]+n-1)+(n-n-1)]+n=2n+2[(n-2)+(n-4)+(n-6)+.0 or 1] plus or minus all added binomial numbers.

and when n is an even number, it is determined by 1+2+3+.n and s = n + (n-1) + (n-2) +1 Sum up:

2s=[n+1]+[n-1)+2]+[n-2)+3]+.n-n-1)+(n-1)]=2[(n-1)+(n-3)+(n-5)+.0 or 1] add or subtract all the added binomial formulas, and combine the two calculations of 2s when n is an even number, and we get s=n+(n-1)+(n-2)+

1.

First, this is a power series, not a power function; Secondly, there is a formula for a power series that the sum function of n from 0 to x is 1 1-x, I don't know if you have heard of it. And the n in your problem starts at 1, which is the formula, removing a first term. And according to the basic properties of the convergence series, it can be determined that the formula of the series I said can be set, and the remaining n sides will find its sum separately, so that it can be proved.

f=∑(t=1,i) at(1+i)^(n-t)=a[ (1+i)^(n-1)+…1+i)^1+(1+i)^0 ]

Using the equation of proportional summation: sn=a1*(1-q n) (1-q) where a1 is the first term and q is the common ratio.

a * 1*[ 1-(1+i)^n ] / [1-(1+i)]=a * 1+i)^n-1]/i

If you don't understand, please ask.