find-the-coefficients-of-x-2-and-x-3-terms-in-the-expansion-of-1-x-1-2x-2-1-3x-3-1-100x-100-

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Question Number 85902 by mr W last updated on 26/Mar/20

$$findthecoefficientsof{x}^{2}and{x}^3$$$$termsintheexpansionof$$$$(1+x){(1+2x)}^2{(1+3x)}^3\dots {(1+100x)}^{100}$$

Commented by Serlea last updated on 26/Mar/20

$$\mathrm{Ok}$$$$\mathrm{From}\mathrm{my}\mathrm{analysis},{\mathrm{it}}^{\prime }\mathrm{s}21868\mathrm{and}1400$$$$\mathbf{How}?$$$$\mathrm{Starting}(1+\mathrm{x})={\mathrm{Doesn}}^{{}^{\prime }}\mathrm{t}\mathrm{exist}$$$$(1+\mathrm{x}){(1+2\mathrm{x})}^2=4\mathrm{and}8$$$$(1+\mathrm{x}){(1+2\mathrm{x})}^2{(1+3\mathrm{x})}^3=238\mathrm{and}80$$$$\{{238\mathrm{x}}^3+{80\mathrm{x}}^2+14\mathrm{x}+1\}$$$$(1+\mathrm{x}){(1+2\mathrm{x})}^2{(1+3\mathrm{x})}^3{(1+4\mathrm{x})}^4=3118\mathrm{and}400$$$$\{{3118\mathrm{x}}^3+{400\mathrm{x}}^2+30\mathrm{x}+1\}$$$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{impossible}\mathrm{to}\mathrm{calculate}\mathrm{to}\mathrm{the}\mathrm{end}\mathbf{B}\mathrm{uy}$$$$\mathrm{let}\mathrm{try}\mathrm{the}\mathrm{next}\mathrm{two}$$$$(1+\mathrm{x}){(1+2\mathrm{x})}^2{(1+3\mathrm{x})}^3{(1+4\mathrm{x})}^4{(1+5\mathrm{x})}^5=21868\mathrm{and}1400$$$$\{{21868\mathrm{x}}^3+{1400\mathrm{x}}^2+55\mathrm{x}+1\}$$$$(1+\mathrm{x}){(1+2\mathrm{x})}^2{(1+3\mathrm{x})}^3{(1+4\mathrm{x})}^4{(1+5\mathrm{x})}^5{(1+6\mathrm{x})}^6$$$$\Rightarrow 21868{(1+6\mathbf{x})}^6{\mathbf{x}}^3+1400{(1+6\mathbf{x})}^6{\mathrm{x}}^2+55{(1+6\mathbf{x})}^6+{(1+6\mathbf{x})}^6$$$$\underset{{21868\mathrm{x}}^3}{\downdownarrows }\underset{{1400\mathrm{x}}^2}{\downdownarrows }$$$$$$$$\mathrm{With}\mathrm{this},\mathrm{The}\mathrm{coeff}.\mathrm{remains}\mathrm{constant}\mathrm{for}$$$$21868\mathrm{and}1400$$$$$$$$$$$$$$

Commented by mr W last updated on 26/Mar/20

$$i{don}^{\prime }tthinkyouarerightsir.$$$$evenwhenweonlyhave{(1+100x)}^{100}$$$$x^{2}termis{C}_2^{100}{\left(100x\right)}^2=\frac{99\times {100}^3}{2}{x}^2$$$$x^{3}termis{C}_3^{100}{\left(100x\right)}^3=\frac{98\times 99\times {100}^4}{6}{x}^3$$

Commented by Serlea last updated on 26/Mar/20

$$\mathrm{Ok}\mathrm{sir}\mathrm{Thanks}\mathrm{for}\mathrm{comment}\mathrm{but}\mathrm{Watch}$$$$\mathrm{The}{\mathrm{c}}_{2,3}^{100}\mathrm{is}\mathrm{going}\mathrm{to}\mathrm{be}\mathrm{multiply}\mathrm{by}\mathrm{another}{\mathrm{X}}^{\mathrm{n}}(\mathrm{n}=\mathrm{certain}\mathrm{number}\ne 0,1)\mathrm{to}\mathrm{increase}\mathrm{the}\mathrm{power}$$$$\frac{98\times 99\times {100}^4}{6}{\mathrm{X}}^{3+\mathrm{n}}$$$$\mathrm{Hope}\mathrm{that}\mathrm{You}\mathrm{review}\mathrm{the}\mathrm{solution}\mathrm{again}$$

Commented by mr W last updated on 26/Mar/20

$$certainlyiknowthat.$$$$ijustwantedtoshowthatyouranswer$$$$iswrong.i{didn}^{\prime }tsaythatisthefinal$$$$answerwhatigave.ijust$$$$showedthe{x}^{2}and{x}^{3}in$$$${(1+100x)}^{100},thecoef.arealready$$$$veryhuge.isaid“evenwhenweonly{\dots }^{″}$$$$theanswertothequestionismuch$$$$morecomplicated.i^{\prime }lltrylater.$$

Commented by Serlea last updated on 26/Mar/20

$$$$$$\mathrm{To}\mathrm{Look}\mathrm{from}\mathrm{the}\mathrm{Open}\mathrm{door},$$$$\mathrm{The}\mathrm{coeff}.\mathrm{Of}{\mathrm{X}}^2\mathrm{and}{\mathrm{X}}^3\mathrm{Will}\mathrm{remain}$$$$\mathrm{constant}\mathrm{because}\mathrm{for}\mathrm{every}\mathrm{Bracket},$$$$\mathrm{There}\mathrm{will}\mathrm{be}\mathrm{a}\mathrm{number}{}^{″}{1}^{″}\mathrm{that}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{a}\mathrm{coeff}.$$$$\mathrm{let}\mathrm{a}=\mathrm{coeff}\mathrm{of}{\mathrm{x}}^3\mathrm{for}{\mathrm{ax}}^3{(1+100\mathrm{x})}^{100}\Rightarrow 1+\dots ..$$$$\mathrm{So}:{\mathrm{ax}}^3(1+12\mathrm{x}+..)$$$$={\mathrm{ax}}^3+\dots ..$$$$\mathrm{That}\mathrm{is}\mathrm{to}\mathrm{show}\mathrm{that}\mathrm{the}\mathrm{coeff}\mathrm{of}{\mathrm{X}}^3$$$$\mathrm{will}\mathrm{remain}\mathrm{constant}$$$$$$

Commented by mr W last updated on 26/Mar/20

$$buteveryxtermandevery{x}^{2}canalso$$$$formanew{x}^{3}term.$$

Commented by Kunal12588 last updated on 26/Mar/20

$$andtheanswersare$$$$x^2\to 57227610000$$$$x^3\to 6451445040986110$$$$I{don}^{\prime }tknowhowtosolve,butthesearethe$$$$answer.$$

Commented by jagoll last updated on 26/Mar/20

$$\mathrm{sir}57227610000={\mathrm{C}}_2^{100}??$$

Commented by naka3546 last updated on 26/Mar/20

$$C\underset{2}{\stackrel{100}{}}=4950$$

Commented by jagoll last updated on 26/Mar/20

$$\mathrm{correction}$$$${\mathrm{C}}_2^{100}\times {\left(100\right)}^2$$

Commented by naka3546 last updated on 26/Mar/20

Commented by Serlea last updated on 26/Mar/20

$$\mathrm{How}\mathrm{is}\mathrm{that}\mathrm{possible}\mathrm{when}\mathrm{it}\mathrm{is}\mathrm{not}$$$${(1+\mathrm{x})}^{100}$$

Commented by TawaTawa1 last updated on 26/Mar/20

$$\mathrm{Sir}\mathrm{mrW}.\mathrm{being}\mathrm{a}\mathrm{while}\mathrm{sir}.$$$$\mathrm{Please}\mathrm{kindly}\mathrm{help}\mathrm{with}\mathrm{question}85950.$$$$\mathrm{Please}.\mathrm{Thanks}\mathrm{for}\mathrm{every}\mathrm{time}\mathrm{sir}.\mathrm{I}\mathrm{need}\mathrm{it}.$$$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by Prithwish Sen 1 last updated on 26/Mar/20

$$\mathrm{Let}$$$$(1+\mathrm{x}){(1+2\mathrm{x})}^2{(1+3\mathrm{x})}^3{(1+4\mathrm{x})}^4$$$$\mathrm{and}\mathrm{for}(1+\mathrm{x}),{\mathrm{A}}^{\mathrm{x}}\mathrm{and}{\mathrm{A}}^{{\mathrm{x}}^2}\mathrm{denotes}\mathrm{the}\mathrm{coeffi}.\mathrm{of}$$$$\mathrm{x}\mathrm{and}{\mathrm{x}}^2\mathrm{repectively}$$$$\mathrm{similarly}\mathrm{B},\mathrm{C},\mathrm{D}\mathrm{for}\mathrm{other}3\mathrm{factors}$$$$\mathrm{then}\mathrm{for}\mathrm{the}\mathrm{coefficient}\mathrm{of}{\mathbf{x}}^2\mathrm{for}\mathrm{the}\mathrm{expression}$$$$\mathrm{will}\mathrm{be}$$$${\mathbf{A}}^{\mathbf{x}}.{\mathbf{B}}^{\mathbf{x}}+{\mathbf{A}}^{\mathbf{x}}.{\mathbf{C}}^{\mathbf{x}}+{\mathbf{A}}^{\mathbf{x}}.{\mathbf{D}}^{\mathbf{x}}+{\mathbf{B}}^{\mathbf{x}}.{\mathbf{C}}^{\mathbf{x}}+{\mathbf{B}}^{\mathbf{x}}.{\mathbf{D}}^{\mathbf{x}}+{\mathbf{C}}^{\mathbf{x}}.{\mathbf{D}}^{\mathbf{x}}+$$$${\mathbf{A}}^{{\mathbf{x}}^2}+{\mathbf{B}}^{{\mathbf{x}}^2}+{\mathbf{C}}^{{\mathbf{x}}^2}$$$$\mathrm{similarly}\mathrm{for}\mathrm{the}\mathrm{coefficient}\mathrm{of}{\mathbf{x}}^3$$$${\mathbf{A}}^{\mathbf{x}}.{\mathbf{B}}^{{\mathbf{x}}^2}+{\mathbf{A}}^{\mathbf{x}}.{\mathbf{C}}^{{\mathbf{x}}^2}+{\mathbf{A}}^{\mathbf{x}}.{\mathbf{D}}^{{\mathbf{x}}^2}+{\mathbf{B}}^{\mathbf{x}}.{\mathbf{C}}^{{\mathbf{x}}^2}+{\mathbf{B}}^{\mathbf{x}}.{\mathbf{D}}^{{\mathbf{x}}^2}+{\mathbf{C}}^{\mathbf{x}}.{\mathbf{D}}^{{\mathbf{x}}^2}$$$$+{\mathbf{B}}^{{\mathbf{x}}^2}.{\mathbf{C}}^{\mathbf{x}}+{\mathbf{B}}^{{\mathbf{x}}^2}.{\mathbf{D}}^{\mathbf{x}}+{\mathbf{C}}^{{\mathbf{x}}^2}.{\mathbf{D}}^{\mathbf{x}}+{\mathbf{C}}^{{\mathbf{x}}^3}+{\mathbf{D}}^{{\mathbf{x}}^3}+{\mathbf{A}}^{\mathbf{x}}.{\mathbf{B}}^{\mathbf{x}}.{\mathbf{C}}^{\mathbf{x}}+{\mathbf{A}}^{\mathbf{x}}.{\mathbf{B}}^{\mathbf{x}}.{\mathbf{D}}^{\mathbf{x}}$$$$+{\mathbf{A}}^{\mathbf{x}}.{\mathbf{C}}^{\mathbf{x}}.{\mathbf{D}}^{\mathbf{x}}+{\mathbf{B}}^{\mathbf{x}}.{\mathbf{C}}^{\mathbf{x}}.{\mathbf{D}}^{\mathbf{x}}$$$$\mathbf{waiting}\mathbf{for}\mathbf{your}\mathbf{response}\mathbf{sir}.$$

Commented by Serlea last updated on 26/Mar/20

$$\mathrm{Can}\mathrm{u}\mathrm{explain}\mathrm{again}$$$$\mathrm{I}\mathrm{am}\mathrm{not}\mathrm{getting}\mathrm{ur}\mathrm{point}$$

Answered by mr W last updated on 06/Apr/20

$$letn=100$$$$P=(1+x){(1+2x)}^2{(1+3x)}^3\dots {(1+100x)}^{100}$$$$=P_1{P}_2{P}_3\dots P_k\dots P_n$$$$P_k={(1+kx)}^k=\underset{r=0}{\sum ^k}{C}_r^k{\left(kx\right)}^r$$$$P_k=1+k\left(kx\right)+\frac{k(k-1)}{2!}{\left(kx\right)}^2+\frac{k(k-1)(k-2)}{3!}{\left(kx\right)}^3+\frac{k(k-1)(k-2)(k-3)}{4!}{\left(kx\right)}^4+o\left(x^4\right)$$$$=1+k^{2}x+\frac{k^3(k-1)}{2}{x}^2+\frac{k^4(k-1)(k-2)}{6}{x}^3+\frac{k^5(k-1)(k-2)(k-3)}{24}{x}^4+o\left(x^4\right)$$$$\Rightarrow P_k=1+a_{k,1}x+a_{k,2}{x}^2+a_{k,3}{x}^3+a_{k,4}{x}^4+o\left(x^4\right)$$$$with$$$$a_{\overline{k},1}=k^2,a_{k,2}=\frac{k^3(k-1)}{2},a_{k,3}=\frac{k^4(k-1)(k-2)}{6},a_{k,4}=\frac{k^5(k-1)(k-2)(k-3)}{24}$$$$k=1,2,3,\dots ,n$$$$$$$$\mathit{c}\mathit{o}\mathit{e}\mathit{f}.\mathit{o}\mathit{f}\mathit{x}\mathit{t}\mathit{e}\mathit{r}\mathit{m}:$$$$C_1=\underset{k=1}{\sum ^n}{a}_{k,1}=\underset{k=1}{\sum ^n}{k}^2=\frac{n(n+1)(2n+1)}{6}$$$$withn=100,$$$$C_1=\frac{100\times 101\times 201}{6}=338350$$$$$$$$\mathit{c}\mathit{o}\mathit{e}\mathit{f}.\mathit{o}\mathit{f}{\mathit{x}}^2\mathit{t}\mathit{e}\mathit{r}\mathit{m}:$$$$C_2=\underset{k=1}{\sum ^n}\underset{j=k+1}{\sum ^n}{a}_{k,1}{a}_{j,1}+\underset{k=1}{\sum ^n}{a}_{k,2}$$$$=\frac{1}{2}\underset{k=1}{\sum ^n}{a}_{k,1}(\underset{j=1}{\sum ^n}{a}_{j,1}-a_{k,1})+\underset{k=1}{\sum ^n}{a}_{k,2}$$$$=\frac{1}{2}[{\left(\underset{k=1}{\sum ^n}{a}_{k,1}\right)}^2-\underset{k=1}{\sum ^n}{a}_{k,1}^2]+\underset{k=1}{\sum ^n}{a}_{\overline{k}2}$$$$=\frac{1}{2}[{\left(\underset{k=1}{\sum ^n}{k}^2\right)}^2-\underset{k=1}{\sum ^n}{k}^4]+\underset{k=1}{\sum ^n}\frac{k^3(k-1)}{2}$$$$=\frac{1}{2}{\left(\underset{k=1}{\sum ^n}{k}^2\right)}^2-\frac{1}{2}\underset{k=1}{\sum ^n}{k}^4+\frac{1}{2}\underset{k=1}{\sum ^n}{k}^4-\frac{1}{2}\underset{k=1}{\sum ^n}{k}^3$$$$=\frac{1}{2}{\left(\underset{k=1}{\sum ^n}{k}^2\right)}^2-\frac{1}{2}\underset{k=1}{\sum ^n}{k}^3$$$$=\frac{1}{2}{\left[\frac{n(n+1)(2n+1)}{6}\right]}^2-\frac{1}{2}{\left[\frac{n(n+1)}{2}\right]}^2$$$$=\frac{(n-1)n^2{(n+1)}^2(n+2)}{18}$$$$withn=100,$$$$C_2=\frac{99\times {100}^2\times {101}^2\times 102}{18}=57227610000$$$$$$$$\left(tobecontinued\right)$$

Commented by TawaTawa1 last updated on 26/Mar/20

$$\mathrm{Sweet},\mathrm{i}\mathrm{will}\mathrm{study}\mathrm{this},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by TawaTawa1 last updated on 26/Mar/20

$$\mathrm{Sir},\mathrm{your}\mathrm{method}\mathrm{too}\mathrm{is}\mathrm{needed}\mathrm{in}\mathrm{Q}85950.$$

Commented by Prithwish Sen 1 last updated on 26/Mar/20

$$\mathrm{excellent}\mathrm{sir}.$$

Commented by Prithwish Sen 1 last updated on 26/Mar/20

$$\mathrm{sir}\mathrm{think}\mathrm{I}\mathrm{get}\mathrm{it}\mathrm{for}{\mathrm{x}}^3.\mathrm{Please}\mathrm{comment}.$$

Commented by Prithwish Sen 1 last updated on 27/Mar/20

$$\mathrm{\Sigma }\mathrm{\Sigma }\mathrm{\Sigma }{\mathrm{a}}_{\mathrm{i},1}{\mathrm{a}}_{\mathrm{j},1}{a}_{\mathrm{k},1}+\mathrm{\Sigma }\mathrm{\Sigma }{\mathrm{a}}_{\mathrm{i},1}{\mathrm{a}}_{\mathrm{j},2}+\mathrm{\Sigma }{\mathrm{a}}_{\mathrm{i},3}$$$$\mathrm{I}\mathrm{think}\mathrm{it}\mathrm{will}\mathrm{be}\mathrm{something}\mathrm{like}\mathrm{this}.$$

Commented by Prithwish Sen 1 last updated on 26/Mar/20

$$\mathrm{I}\mathrm{think}\mathrm{u}\mathrm{bave}\mathrm{missed}\mathrm{the}\mathrm{product}\mathrm{of}$$$${\mathrm{a}}_{\mathrm{i},1}.{\mathrm{a}}_{\mathrm{j},1}.{\mathrm{a}}_{\mathrm{k},1}$$

Commented by mr W last updated on 05/Apr/20

$$\left(continued\right)$$$$$$$$let$$$$K_2=\underset{k=1}{\sum ^n}{k}^2=\frac{n(n+1)(2n+1)}{6}$$$$K_3=\underset{k=1}{\sum ^n}{k}^3=\frac{n^2{(n+1)}^2}{4}$$$$K_4=\underset{k=1}{\sum ^n}{k}^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}=\frac{3n^2+3n-1}{5}{K}_2$$$$K_5=\underset{k=1}{\sum ^n}{k}^5=\frac{n^2{(n+1)}^2(2n^2+2n-1)}{12}=\frac{2n^2+2n-1}{3}{K}_3$$$$K_6=\underset{k=1}{\sum ^n}{k}^6=\frac{n(n+1)(2n+1)(3n^4+6n^3-3n+1)}{42}=\frac{3n^4+6n^3-3n+1}{7}{K}_2$$$$$$$$\mathit{c}\mathit{o}\mathit{e}\mathit{f}.\mathit{o}\mathit{f}{\mathit{x}}^3\mathit{t}\mathit{e}\mathit{r}\mathit{m}:$$$$C_3=C_{31}+C_{32}+C_{33}$$$$with$$$$C_{31}=\underset{k\ne j\ne i}{\sum ^n}{a}_{k,1}{a}_{j,1}{a}_{i,1}$$$$C_{32}=\underset{k,j=1,j\ne k}{\sum ^n}{a}_{k,1}{a}_{j,2}$$$$C_{33}=\underset{k=1}{\sum ^n}{a}_{k,3}$$$$$$$$C_{31}=\frac{1}{6}[{\left(\underset{k=1}{\sum ^n}{a}_{k,1}\right)}^3-\underset{k=1}{\sum ^n}{a}_{k,1}^3-3\underset{i=1}{\sum ^n}{a}_{i,1}^2(\underset{k=1}{\sum ^n}{a}_{k,1}-a_{i,1})]$$$$=\frac{1}{6}[{\left(\underset{k=1}{\sum ^n}{a}_{k,1}\right)}^3-\underset{k=1}{\sum ^n}{a}_{k,1}^3-3\left(\underset{i=1}{\sum ^n}{a}_{i,1}^2\right)\left(\underset{k=1}{\sum ^n}{a}_{k,1}\right)+3\underset{i=1}{\sum ^n}{a}_{i,1}^3]$$$$=\frac{1}{6}(K_2^3-3K_4{K}_2+2K_6)$$$$$$$$C_{32}=\underset{k=1}{\sum ^n}{a}_{k,1}\underset{j=1,j\ne k}{\sum ^n}{a}_{j,2}$$$$=\underset{k=1}{\sum ^n}{a}_{k,1}(\underset{j=1}{\sum ^n}{a}_{j,2}-a_{k,2})$$$$=\left(\underset{k=1}{\sum ^n}{a}_{k,1}\right)\left(\underset{j=1}{\sum ^n}{a}_{j,2}\right)-\underset{k=1}{\sum ^n}{a}_{k,1}{a}_{k,2}$$$$=\left(\underset{k=1}{\sum ^n}{k}^2\right)\left(\underset{k=1}{\sum ^n}\frac{k^3(k-1)}{2}\right)-\underset{k=1}{\sum ^n}\frac{k^2{k}^3(k-1)}{2}$$$$=\frac{1}{2}(K_2{K}_4-K_2{K}_3-K_6+K_5)$$$$$$$$C_{33}=\underset{k=1}{\sum ^n}{a}_{k,3}$$$$=\underset{k=1}{\sum ^n}\frac{k^4(k-1)(k-2)}{6}$$$$=\frac{1}{6}(K_6-3K_5+2K_4)$$$$$$$$C_3=\frac{1}{6}(K_2^3-3K_2{K}_4+2K_6)+\frac{1}{2}(K_2{K}_4-K_2{K}_3-K_6+K_5)+\frac{1}{6}(K_6-3K_5+2K_4)$$$$=\frac{1}{6}(K_2^3-3K_2{K}_3+2K_4)$$$$=\frac{n(n+1)(2n+1)}{36}[\frac{n^2{(n+1)}^2{(2n+1)}^2}{36}-\frac{3n^2{(n+1)}^2}{4}+\frac{2(3n^2+3n-1)}{5}]$$$$=\frac{n(n+1)(2n+1)}{36}[\frac{n^2{(n+1)}^2(2n^2+2n-13)}{18}+\frac{2(3n^2+3n-1)}{5}]$$$$=\frac{(n-1)n(n+1)(n+2)(2n+1)(10n^4+20n^3-35n^2-45n+18)}{3240}$$$$withn=100,$$$$C_3=\frac{99\times 100\times 101\times 102\times 201\times (10\times {100}^4+20\times {10}^3-35\times {100}^2-45\times 100+18)}{3240}$$$$=6451445040986110$$

Commented by Prithwish Sen 1 last updated on 27/Mar/20

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{fix}\mathrm{it}\mathrm{sir}.$$

Commented by mr W last updated on 06/Apr/20

$$\mathit{c}\mathit{o}\mathit{e}\mathit{f}.\mathit{o}\mathit{f}{\mathit{x}}^4\mathit{t}\mathit{e}\mathit{r}\mathit{m}:$$$$$$$$C_4=C_{41}+C_{42}+C_{43}+C_{44}+C_{45}$$$$with$$$$C_{41}=\sum _{k\ne i\ne j\ne h}{a}_{k,1}{a}_{i,1}{a}_{j,1}{a}_{h,1}$$$$C_{42}=\sum _{k\ne i\ne j}{a}_{k,1}{a}_{i,1}{a}_{j,2}$$$$C_{43}=\sum _{k\ne i}{a}_{k,1}{a}_{i,3}$$$$C_{44}=\sum _{k\ne i}{a}_{k,2}{a}_{i,2}$$$$C_{45}=\underset{k=1}{\sum ^n}{a}_{k,4}$$$$\dots ..$$$$C_{42}=\sum _{k\ne i\ne j}{a}_{k,1}{a}_{i,1}{a}_{j,2}$$$$$$$$$$$$C_{43}=\sum _{k\ne i}{a}_{k,1}{a}_{i,3}$$$$=\frac{1}{2}\underset{k=1}{\sum ^n}{a}_{k,1}(\underset{i=1}{\sum ^n}{a}_{i,3}-a_{k,3})$$$$=\frac{1}{2}[\left(\underset{k=1}{\sum ^n}{a}_{k,1}\right)\left(\underset{i=1}{\sum ^n}{a}_{i,3}\right)-\underset{k=1}{\sum ^n}{a}_{k,1}{a}_{k,3}]$$$$=\frac{1}{2}[\left(\underset{k=1}{\sum ^n}{k}^2\right)\left(\underset{i=1}{\sum ^n}\frac{k^4(k-1)(k-2)}{6}\right)-\underset{k=1}{\sum ^n}\frac{k^6(k-1))k-2)}{6}]$$$$=\frac{1}{12}[\left(\underset{k=1}{\sum ^n}{k}^2\right)\left(\underset{i=1}{\sum ^n}{k}^4(k-1)(k-2)\right)-\underset{k=1}{\sum ^n}{k}^6(k-1)(k-2)]$$$$=\frac{1}{12}[\left(\underset{k=1}{\sum ^n}{k}^2\right)\left(\underset{i=1}{\sum ^n}(k^6-3k^5+2k^4)\right)-\underset{k=1}{\sum ^n}(k^8-3k^7+2k^6)]$$$$=\frac{1}{12}[K_2(K_6-3K_5+2K_4)-(K_8-3K_7+2K_6)]$$$$=\frac{1}{12}(K_2{K}_6-3K_2{K}_5+2K_2{K}_4-K_8+3K_7-2K_6)$$$$$$$$C_{44}=\sum _{k\ne i}{a}_{k,2}{a}_{i,2}$$$$=\frac{1}{2}\underset{k=1}{\sum ^n}{a}_{k,2}(\underset{i=1}{\sum ^n}{a}_{i,2}-a_{k,2})$$$$=\frac{1}{2}[{\left(\underset{k=1}{\sum ^n}{a}_{k,2}\right)}^2-\underset{k=1}{\sum ^n}{a}_{k,2}^2]$$$$=\frac{1}{2}[{\left(\underset{k=1}{\sum ^n}\frac{k^3(k-1)}{2}\right)}^2-\underset{k=1}{\sum ^n}\frac{k^6{(k-1)}^2}{4}]$$$$=\frac{1}{8}[{(K_4-K_3)}^2-(K_8-2K_7+K_6)]$$$$=\frac{1}{8}(K_4^2-2K_3{K}_4+K_3^2-K_8+2K_7-K_6)$$$$$$$$C_{45}=\underset{k=1}{\sum ^n}{a}_{k,4}=\underset{k=1}{\sum ^n}\frac{k^5(k-1)(k-2)(k-3)}{24}$$$$=\frac{1}{24}(K_8-6K_7+11K_6-6K_5)$$$$\dots \dots$$

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