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Question Number 46186 by Saorey last updated on 22/Oct/18

$$\mathrm{please}\mathrm{help}\mathrm{me}!$$$$\mathrm{S}=1^2{\mathrm{q}}^1+2^2{\mathrm{q}}^2+3^2{\mathrm{q}}^3+...+{\mathrm{n}}^2{\mathrm{q}}^{\mathrm{n}}=?$$

Commented by maxmathsup by imad last updated on 22/Oct/18

$$wehaveS=\sum _{p=0}^n{p}^2{q}^{p}letp\left(x\right)=\sum _{p=0}^n{x}^{p}withx\ne 1\Rightarrow$$$$p^{{}^{\prime }}\left(x\right)=\sum _{p=1}^{n}p{x}^{p-1}\Rightarrow x{p}^{{}^{\prime }}\left(x\right)=\sum _{p=1}^n{px}^p\Rightarrow p^{{}^{\prime }}\left(x\right)+x{p}^{{}^{″}}\left(x\right)=\sum _{p=1}^n{p}^2{x}^{p-1}\Rightarrow$$$${xp}^{{}^{\prime }}\left(x\right)+x^2{p}^{{}^{″}}\left(x\right)=\sum _{p=1}^n{p}^2{x}^{p}butp\left(x\right)=\frac{x^{n+1}-1}{x-1}\Rightarrow$$$$p^{{}^{\prime }}\left(x\right)=\frac{{nx}^{n+1}-(n+1)x^n+1}{{(x-1)}^2}and$$$$p^{\left(2\right)}\left(x\right)=\frac{(n(n+1)x^n-n(n+1)x^{n-1}){(x-1)}^2-2(x-1)({nx}^{n+1}-(n+1)x^n+1)}{{(x-1)}^4}$$$$=\frac{(n^2+n)(x^n-x^{n-1})(x-1)-2n{x}^{n+1}+2(n+1)x^n-2}{{(x-1)}^3}$$$$=\frac{(n^2+n)(x^{n+1}-2x^n+x^{n-1)})-2n{x}^{n+1}+2(n+1)x^n-2}{{(x-1)}^3}$$$$=\frac{(n^2-n)x^{n+1}+(-2n^2-2n+2n+2)x^n+(n^2+n)x^{n-1}-2}{{(x-1)}^3}$$$$=\frac{(n^2-n)x^{n+1}-2(n^2-1)x^n+(n^2+n)x^{n-1}-2}{{(x-1)}^3}\Rightarrow$$$$S=\sum _{p=1}^n{p}^2{q}^p={qp}^{{}^{\prime }}\left(q\right)+q^2{p}^{{}^{″}}\left(q\right)$$$$=\frac{q}{{(q-1)}^2}\{{nq}^{n+1}-(n+1)q^n+1\}+\frac{q^2}{{(q-1)}^3}\{(n^2-n)q^{n+1}-2(n^2-1)q^n+(n^2+n)q^{n-1}-2\}if$$$$q\ne 1andifq=1S=1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}.$$$$$$

Answered by ajfour last updated on 22/Oct/18

$$\frac{S}{q}=1+2^{2}q+3^2{q}^2+....+n^2{q}^{n-1}$$$$S(\frac{1}{q}-1)=1+3q+5q^2+7q^3+...$$$$....+(2n-1)q^{n-1}-n^2{q}^n$$$$Sq(\frac{1}{q}-1)=q+3q^2+5q^3+...$$$$....+(2n-1)q^n-n^2{q}^{n+1}$$$$S(\frac{1}{q}-1)(1-q)=1+2q+2q^2+2q^3+..$$$$..+2q^{n-1}-{(n-1)}^2{q}^n+n^2{q}^{n+1}$$$$\Rightarrow \frac{{(1-q)}^{2}S}{q}=1+\frac{2q(1-q^n)}{1-q}$$$$-{(n-1)}^2{q}^n+n^2{q}^{n+1}$$$$\mathit{S}=\frac{\mathit{q}}{{(1-\mathit{q})}^2}[1+\frac{2\mathit{q}(1-{\mathit{q}}^{\mathit{n}})}{1-\mathit{q}}$$$$-{(\mathit{n}-1)}^2{\mathit{q}}^{\mathit{n}}+{\mathit{n}}^2{\mathit{q}}^{\mathit{n}+1}].$$$$(thereislittleerror,ishall$$$$soonfixit..)$$

Commented by Saorey last updated on 22/Oct/18

$$\mathrm{thank}\mathrm{you}$$

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