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Question Number 147791 by Tawa11 last updated on 23/Jul/21

$$\mathrm{If}\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=1$$$${\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+{\mathrm{d}}^2=2$$$${\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3+{\mathrm{d}}^3=3$$$${\mathrm{a}}^4+{\mathrm{b}}^4+{\mathrm{c}}^4+{\mathrm{d}}^4=4$$$$\mathrm{Evaluate}:$$$${\mathrm{a}}^6+{\mathrm{b}}^6+{\mathrm{c}}^6+{\mathrm{d}}^6$$

Commented by prakash jain last updated on 25/Jul/21

$$g\left(k\right)=\frac{1}{1-ka}=1+ka+k^2{a}^2+..$$$$f\left(k\right)=\frac{1}{1-ka}+\frac{1}{1-kb}+\frac{1}{1-kc}+\frac{1}{1-kd}$$$$f\left(k\right)=\frac{4+A_{1}k+A_2{k}^2+A_3{k}^3}{1+B_{1}k+B_2{k}^2+B_3{k}^3+B_4{k}^2}$$$$A_1=-3(a+b+c+d)$$$$B_1=-(a+b+c+d)\Rightarrow A_1=3B_1$$$$A_2=2(ab+bc+ad+bc+bd+cd)=2B_2$$$$A_3=-(abc+abd+acd+bcd)=B_3$$$$f\left(k\right)=4+k(a+b+c+d)+k^2(a^2+b^2+c^2+d^2)..$$$$\mathrm{Let}\mathrm{us}\mathrm{denote}{a}^n+b^n+c^n+d^n=u_n$$$$f\left(k\right)=4+{ku}_1+k^2{u}_2+k^3{u}_3+k^4{u}_4+...$$$$4+A_{1}k+A_2{k}^2+A_3{k}^3=$$$$(1+B_{1}k+B_2{k}^2+B_3{k}^3)(4+{ku}_1+k^2{u}_2+k^3{u}_3+k^4{u}_4+k^5{u}_5+k^{6}u6+...)$$$$\mathrm{Given}{u}_1=1,u_2=2,u_3=3,u_4=4$$$$\mathrm{C}omparingcoefficients$$$$A_1=u_1+4B_1\Rightarrow 3B_1=1+4B_1$$$$\Rightarrow B_1=-1,A_1=-3$$$$A_2=u_2+u_1{B}_1+4B_2=2-1+4B_2$$$$\Rightarrow 2B_2=1+4B_2\Rightarrow B_2=\frac{-1}{2},A_2=-1$$$$A_3=u_3+B_1{u}_2+B_2{u}_1+4B_3$$$$\Rightarrow B_3=3-2-\frac{1}{2}+4B_3\Rightarrow B_3=A_3=-\frac{1}{6}$$$$0=u_4+B_1{u}_3+B_2{u}_2+B_3{u}_1+4B_4$$$$\Rightarrow 0=4-3-\frac{1}{2}\times 2-\frac{1}{6}+4B_4\Rightarrow B_4=\frac{1}{24}$$$$0=u_5+B_1{u}_4+B_2{u}_3+B_3{u}_2+B_4{u}_1$$$$u_1..u_4,B_1...B_4\mathrm{are}\mathrm{known}{u}_5\mathrm{can}\mathrm{be}$$$$\mathrm{calculated}$$$$\mathrm{similary}$$$$0=u_6+B_1{u}_5+B_2{u}_4+B_3{u}_3+B_4{u}_2$$$$u_1...u_5\mathrm{are}\mathrm{known}\mathrm{and}\mathrm{you}\mathrm{can}$$$$\mathrm{calculate}{u}_6=a^6+b^6+c^6+d^6$$$$$$

Commented by prakash jain last updated on 25/Jul/21

$$u_5-4-\frac{3}{2}-\frac{1}{3}+\frac{1}{24}=0$$$$u_5=\frac{96+36+8-1}{24}=\frac{139}{24}$$$$u_6-\frac{139}{24}-\frac{1}{2}\times 4-\frac{1}{6}\times 3+\frac{1}{24}\times 2$$$$u_6=\frac{139}{24}+2+\frac{1}{2}-\frac{1}{12}=\frac{139+48+12-2}{24}$$$$=\frac{197}{24}$$$$a^6+b^6+c^6+d^6=\frac{197}{24}$$$$$$$$\mathrm{I}{\mathrm{haven}}^{\prime }\mathrm{t}\mathrm{checked}\mathrm{numerical}\mathrm{calculation}$$$$\mathrm{thoroughly}\mathrm{so}\mathrm{please}\mathrm{recheck}\mathrm{these}.$$$$\mathrm{You}\mathrm{can}\mathrm{use}\mathrm{this}\mathrm{method}\mathrm{to}\mathrm{find}$$$$\mathrm{any}{u}_n=a^n+b^n+c^n+d^n$$

Commented by Tawa11 last updated on 25/Jul/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Answered by Rasheed.Sindhi last updated on 23/Jul/21

$$Let\mathrm{b}=\mathrm{pa},\mathrm{c}=\mathrm{qa},\mathrm{d}=\mathrm{ra}$$$$\left(\mathrm{i}\right)\Rightarrow \mathrm{a}(1+\mathrm{p}+\mathrm{q}+\mathrm{r})=1$$$$\Rightarrow \mathrm{p}+\mathrm{q}+\mathrm{r}=1/\mathrm{a}-1=(1-\mathrm{a})/\mathrm{a}...A$$$$\left(\mathrm{ii}\right)\Rightarrow {\mathrm{a}}^2(1+{\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2)=2$$$$\Rightarrow {\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2=2/{\mathrm{a}}^2-1=(2-{\mathrm{a}}^2)/{\mathrm{a}}^2...B$$$$\left(\mathrm{iii}\right)\Rightarrow {\mathrm{p}}^3+{\mathrm{q}}^3+{\mathrm{r}}^3=(3-{\mathrm{a}}^3)/{\mathrm{a}}^3...C$$$$\left(\mathrm{iv}\right)\Rightarrow {\mathrm{p}}^4+{\mathrm{q}}^4+{\mathrm{r}}^4=(4-{\mathrm{a}}^4)/{\mathrm{a}}^4...D$$$$A\Rightarrow {(\mathrm{p}+\mathrm{q}+\mathrm{r})}^2={\left\{(1-\mathrm{a})/\mathrm{a}\right\}}^2$$$${\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2+2(\mathrm{pq}+\mathrm{qr}+\mathrm{rp})={\left(\frac{1-\mathrm{a}}{\mathrm{a}}\right)}^2$$$$\frac{2-{\mathrm{a}}^2}{{\mathrm{a}}^2}+2(\mathrm{pq}+\mathrm{qr}+\mathrm{rp})={\left(\frac{1-\mathrm{a}}{\mathrm{a}}\right)}^2$$$$2(\mathrm{pq}+\mathrm{qr}+\mathrm{rp})={\left(\frac{1-\mathrm{a}}{\mathrm{a}}\right)}^2-\frac{2-{\mathrm{a}}^2}{{\mathrm{a}}^2}$$$$\mathrm{pq}+\mathrm{qr}+\mathrm{rp}=\frac{{(1-\mathrm{a})}^2-2+{\mathrm{a}}^2}{{2\mathrm{a}}^2}$$$$\mathrm{C}\Rightarrow {\mathrm{p}}^3+{\mathrm{q}}^3+{\mathrm{r}}^3=\frac{3-{\mathrm{a}}^3}{{\mathrm{a}}^3}$$$${\mathrm{p}}^3+{\mathrm{q}}^3+{\mathrm{r}}^3-3\mathrm{pqr}=\frac{3-{\mathrm{a}}^3}{{\mathrm{a}}^3}-3\mathrm{pqr}$$$$(\mathrm{p}+\mathrm{q}+\mathrm{r})({\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2-(\mathrm{pq}+\mathrm{qr}+\mathrm{rp}))$$$$=\frac{3-{\mathrm{a}}^3}{{\mathrm{a}}^3}-3\mathrm{pqr}$$$$3\mathrm{pqr}=\frac{3-{\mathrm{a}}^3}{{\mathrm{a}}^3}-\left(\frac{1-\mathrm{a}}{\mathrm{a}}\right)(\frac{2-{\mathrm{a}}^2}{{\mathrm{a}}^2}-\frac{{(1-\mathrm{a})}^2-2+{\mathrm{a}}^2}{{2\mathrm{a}}^2})$$$$\mathrm{pqr}=\frac{3-{\mathrm{a}}^3}{{3\mathrm{a}}^3}-\left(\frac{1-\mathrm{a}}{3\mathrm{a}}\right)(\frac{2-{\mathrm{a}}^2}{{\mathrm{a}}^2}-\frac{{(1-\mathrm{a})}^2-2+{\mathrm{a}}^2}{{2\mathrm{a}}^2})$$

Commented by Tawa11 last updated on 23/Jul/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Answered by mr W last updated on 23/Jul/21

$$p_k=a^k+b^k+c^k+d^k$$$$e_1=p_1=1$$$$2e_2=e_1{p}_1-p_2=1-2=-1\Rightarrow e_2=-\frac{1}{2}$$$$3e_3=e_2{p}_1-e_1{p}_2+p_3=-\frac{1}{2}-2+3=\frac{1}{2}\Rightarrow e_3=\frac{1}{6}$$$$4e_4=e_3{p}_1-e_2{p}_2+e_1{p}_3-p_4=\frac{1}{6}+1+3-4=\frac{1}{6}\Rightarrow e_4=\frac{1}{24}$$$$0=e_4{p}_1-e_3{p}_2+e_2{p}_3-e_1{p}_4+p_5$$$$\Rightarrow p_5=-\frac{1}{24}+\frac{1}{3}+\frac{3}{2}+4=\frac{139}{24}$$$$0=-e_4{p}_2+e_3{p}_3-e_2{p}_4+e_1{p}_5-p_6$$$$\Rightarrow p_6=-\frac{1}{12}+\frac{1}{2}+2+\frac{139}{24}=\frac{197}{24}$$$$$$$$generally$$$$p_n=p_{n-1}+\frac{p_{n-2}}{2}+\frac{p_{n-3}}{6}-\frac{p_{n-4}}{24}$$$$or$$$$p_n=a^n+b^n+c^n+d^n$$$$a,b,c,darerootsofequation$$$$x^4-x^3-\frac{1}{2}{x}^2-\frac{1}{6}x+\frac{1}{24}=0$$

Commented by Tawa11 last updated on 23/Jul/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Commented by mr W last updated on 23/Jul/21

https://en.m.wikipedia.org/wiki/Newton%27s_identities

Commented by Rasheed.Sindhi last updated on 23/Jul/21

$$\mathcal{W}onderfulmethod\mathit{s}\mathit{i}\mathit{r}!$$$$\mathcal{B}ut{I}^{\prime }mfeelingdifficulityin$$$$understanding.$$

Commented by mr W last updated on 23/Jul/21

$$seealsoQ74970$$

Commented by Tawa11 last updated on 23/Jul/21

$$\mathrm{Sir},\mathrm{which}\mathrm{sequence}\mathrm{you}\mathrm{use}\mathrm{to}\mathrm{get}\mathrm{the}{\mathrm{P}}_{\mathrm{n}}.\mathrm{Just}\mathrm{the}\mathrm{sequence}.$$

Commented by Tawa11 last updated on 23/Jul/21

$$\mathrm{Or}\mathrm{there}\mathrm{is}\mathrm{a}\mathrm{format}\mathrm{to}\mathrm{get}\mathrm{the}{\mathrm{P}}_{\mathrm{n}}$$

Commented by Tawa11 last updated on 24/Jul/21

$$\mathrm{I}\mathrm{grab}\mathrm{now},\mathrm{you}\mathrm{used}{\mathrm{e}}_1-{\mathrm{e}}_2+{\mathrm{e}}_3-{\mathrm{e}}_4$$

Commented by peter frank last updated on 24/Jul/21

$$thankyou$$

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