math_ Summation and derivation of proportional series & derivation of sum and difference of equal powers / difference between two nth power numbers/

math_ Sum and derivation of the sequence of proportional numbers & Derivation of sum and difference of equal powers / Two n Power difference

Two n Power difference & and

$$a^n-b^n=(a-b)\sum _{i=0}^{n-1}{a}^{n-1-i}{b}^i$$

among ,a Index and b The sum of the indices is n-1

• It is possible to obtain the deformation of the summation formula through the sequence of proportional numbers ( Dislocation subtraction )

Derivation of the summation formula of the equal ratio sequence

• The sequence $x_n=a_1,a_2,a_3,\dots ,a_n$(n term )

• General formula $x_n=a_1{q}^{n-1},(n=1,2,3,\dots ,n-1,n),branchothersurfaceinCountColumnOfThe\; first1,2,3,...nterm$

• front n Xiang He :

  • $$\begin{gathered} S_n=a_1+a_2+a_3+\cdots +a_{n-1}+a_n \\ =a_1{q}^0+a_1{q}^1+\cdots +a_1{q}^{n-1} \\ =a_1(q^0+q^1+\cdots +q^{n-1})=a_1\sum _{i=0}^{i=n-1}{q}^i \end{gathered}$$

  • Special , When a1=0 when ,

  • $$\begin{gathered} S_n=a_1\sum _{i=0}^{i=n-1}{q}^i=1\times \sum _{i=0}^{i=n-1}{q}^i=q^0+q^1+q^2+\cdots +q^{n-1} \\ canWithuseOnPUSHguideetc.powerandBadMaletype \end{gathered}$$

• In order to adopt Subtraction by dislocation solve , We also need $q{S}_n$,$q{S}_n=a_1(q^1+q^2+\cdots +q^n)=a_1\sum _{i=1}^{i=n}{q}^i$

• $$\begin{gathered} q{S}_n-S_n \\ =a_1\sum _{i=1}^{i=n}{q}^i-a_1\sum _{i=0}^{i=n-1}{q}^i \\ =a_1((\sum _{i=1}^{i=n-1}{q}^i+q^n)-(q^0+\sum _{i=1}^{i=n-1}{q}^i)) \\ =a_1(q^n-q^0)=a_1(q^n-1) \end{gathered}$$

$$\begin{gathered} S_n(q-1)=a_1(q^n-1) \\ {S}_n=\frac{a_1(q^n-1)}{q-1} \\ ORanotherform: \\ {S}_n=\frac{a_1(1-q^n)}{1-q} \end{gathered}$$

Derivation of equal power difference formula

• Next, we prove the formula of equal power difference

• It is known from the previous derivation that :

  • Current first item a1=1 When , The isometric sequence degenerates into a simpler form :

• $$\begin{gathered} S_n=\sum _{i=0}^{i=n-1}{q}^i=q^0+q^1+q^2+\cdots +q^{n-1} \\ =\frac{q^n-1}{q-1} \end{gathered}$$

• Make $q=\frac{a}{b}$, Bring this formula into the above equation q:

  • $$S_n=\frac{q^n-1}{q-1}=\frac{a^n-b^n}{b^{n-1}(a-b)}$$

    $$\begin{gathered} S_n=\sum _{i=0}^{i=n-1}{q}^i=\sum _{i=0}^{i=n-1}\frac{a^i}{b^i} \\ =\sum _{i=0}^{i=n-1}{a}^i{b}^{-i} \\ {b}^{n-1}{S}_n=b^{n-1}\sum _{i=0}^{i=n-1}{a}^i{b}^{-i}=\sum _{i=0}^{i=n-1}{a}^i{b}^{-i}{b}^{n-1}=\sum _{i=0}^{i=n-1}{a}^i{b}^{n-1-i} \\ {a}^n-b^n=b^{n-1}{S}_n(a-b) \\ =(a-b)\sum _{i=0}^{i=n-1}{a}^i{b}^{n-1-i} \\ fromseekandorderColumnOfYescallsex: \\ =(a-b)\sum _{i=0}^{i=n-1}{a}^{n-1-i}{b}^i \\ benefituseaOffingerCountandbOffingerCountandbyc \\ =(a-b)\sum _{i=1}^{i=n}{a}^{n-i}{b}^{i-1} \end{gathered}$$

• Sum and difference of equal powers - Wikipedia , An encyclopedia of freedom (wikipedia.org)