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

please help me! S=1^2 q^1 +2^2 q^2 +3^2 q^3 +...+n^2 q^n =?

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

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

we have S =Σ_(p=0) ^n p^2 q^p let p(x)=Σ_(p=0) ^n x^p with x≠1 ⇒ p^′ (x)=Σ_(p=1) ^n p x^(p−1) ⇒x p^′ (x)=Σ_(p=1) ^n px^p ⇒p^′ (x)+x p^(′′) (x)=Σ_(p=1) ^n p^2 x^(p−1) ⇒ xp^′ (x) +x^2 p^(′′) (x)=Σ_(p=1) ^n p^2 x^p but p(x)=((x^(n+1) −1)/(x−1)) ⇒ p^′ (x)=((nx^(n+1) −(n+1)x^n +1)/((x−1)^2 )) and p^((2)) (x) =(((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 )) =(((n^2 +n)(x^n −x^(n−1) )(x−1)−2n x^(n+1) +2(n+1)x^n −2)/((x−1)^3 )) =(((n^2 +n)(x^(n+1) −2x^n +x^(n−1)) )−2n x^(n+1) +2(n+1)x^n −2)/((x−1)^3 )) =(((n^2 −n)x^(n+1) +(−2n^2 −2n +2n+2)x^n +(n^2 +n)x^(n−1) −2)/((x−1)^3 )) =(((n^2 −n)x^(n+1) −2(n^2 −1)x^n +(n^2 +n)x^(n−1) −2)/((x−1)^3 )) ⇒ S =Σ_(p=1) ^n p^2 q^p =qp^′ (q)+q^2 p^(′′) (q) =(q/((q−1)^2 )){nq^(n+1) −(n+1)q^n +1} +(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≠1 and if q=1 S =1^2 +2^2 +3^2 +...+n^2 =((n(n+1)(2n+1))/6) .

$${we}\:{have}\:{S}\:=\sum_{{p}=\mathrm{0}} ^{{n}} \:{p}^{\mathrm{2}} {q}^{{p}} \:\:\:\:{let}\:{p}\left({x}\right)=\sum_{{p}=\mathrm{0}} ^{{n}} \:{x}^{{p}} \:{with}\:{x}\neq\mathrm{1}\:\Rightarrow \\ $$$${p}^{'} \left({x}\right)=\sum_{{p}=\mathrm{1}} ^{{n}} \:{p}\:{x}^{{p}−\mathrm{1}} \:\Rightarrow{x}\:{p}^{'} \left({x}\right)=\sum_{{p}=\mathrm{1}} ^{{n}} \:{px}^{{p}} \:\Rightarrow{p}^{'} \left({x}\right)+{x}\:{p}^{''} \left({x}\right)=\sum_{{p}=\mathrm{1}} ^{{n}} {p}^{\mathrm{2}} {x}^{{p}−\mathrm{1}} \:\Rightarrow \\ $$$${xp}^{'} \left({x}\right)\:+{x}^{\mathrm{2}} {p}^{''} \left({x}\right)=\sum_{{p}=\mathrm{1}} ^{{n}} {p}^{\mathrm{2}} {x}^{{p}} \:\:{but}\:{p}\left({x}\right)=\frac{{x}^{{n}+\mathrm{1}} −\mathrm{1}}{{x}−\mathrm{1}}\:\Rightarrow \\ $$$${p}^{'} \left({x}\right)=\frac{{nx}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){x}^{{n}} \:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:{and}\: \\ $$$${p}^{\left(\mathrm{2}\right)} \left({x}\right)\:=\frac{\left({n}\left({n}+\mathrm{1}\right){x}^{{n}} −{n}\left({n}+\mathrm{1}\right){x}^{{n}−\mathrm{1}} \right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{2}\left({x}−\mathrm{1}\right)\left({nx}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){x}^{{n}} \:+\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\frac{\left({n}^{\mathrm{2}} +{n}\right)\left({x}^{{n}} −{x}^{{n}−\mathrm{1}} \right)\left({x}−\mathrm{1}\right)−\mathrm{2}{n}\:{x}^{{n}+\mathrm{1}} +\mathrm{2}\left({n}+\mathrm{1}\right){x}^{{n}} −\mathrm{2}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left({n}^{\mathrm{2}} \:+{n}\right)\left({x}^{{n}+\mathrm{1}} −\mathrm{2}{x}^{{n}} \:+{x}^{\left.{n}−\mathrm{1}\right)} \right)−\mathrm{2}{n}\:{x}^{{n}+\mathrm{1}} +\mathrm{2}\left({n}+\mathrm{1}\right){x}^{{n}} −\mathrm{2}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left({n}^{\mathrm{2}} −{n}\right){x}^{{n}+\mathrm{1}} \:+\left(−\mathrm{2}{n}^{\mathrm{2}} −\mathrm{2}{n}\:+\mathrm{2}{n}+\mathrm{2}\right){x}^{{n}} \:+\left({n}^{\mathrm{2}} \:+{n}\right){x}^{{n}−\mathrm{1}} −\mathrm{2}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left({n}^{\mathrm{2}} −{n}\right){x}^{{n}+\mathrm{1}} \:−\mathrm{2}\left({n}^{\mathrm{2}} −\mathrm{1}\right){x}^{{n}} \:+\left({n}^{\mathrm{2}} \:+{n}\right){x}^{{n}−\mathrm{1}} −\mathrm{2}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$${S}\:=\sum_{{p}=\mathrm{1}} ^{{n}} {p}^{\mathrm{2}} {q}^{{p}} \:={qp}^{'} \left({q}\right)+{q}^{\mathrm{2}} {p}^{''} \left({q}\right) \\ $$$$=\frac{{q}}{\left({q}−\mathrm{1}\right)^{\mathrm{2}} }\left\{{nq}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){q}^{{n}} \:+\mathrm{1}\right\}\:+\frac{{q}^{\mathrm{2}} }{\left({q}−\mathrm{1}\right)^{\mathrm{3}} }\left\{\left({n}^{\mathrm{2}} −{n}\right){q}^{{n}+\mathrm{1}} −\mathrm{2}\left({n}^{\mathrm{2}} −\mathrm{1}\right){q}^{{n}} +\left({n}^{\mathrm{2}} \:+{n}\right){q}^{{n}−\mathrm{1}} −\mathrm{2}\right\}\:{if} \\ $$$${q}\neq\mathrm{1}\:{and}\:{if}\:{q}=\mathrm{1}\:\:{S}\:=\mathrm{1}^{\mathrm{2}} \:+\mathrm{2}^{\mathrm{2}} \:+\mathrm{3}^{\mathrm{2}} \:+...+{n}^{\mathrm{2}} =\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:. \\ $$$$ \\ $$

Answered by ajfour last updated on 22/Oct/18

(S/q)= 1+2^2 q+3^2 q^2 +....+n^2 q^( n−1) S((1/q)−1)= 1+3q+5q^2 +7q^3 +... ....+(2n−1)q^( n−1) −n^2 q^n Sq((1/q)−1)=q+3q^2 +5q^3 +... ....+(2n−1)q^( n) −n^2 q^(n+1) S((1/q)−1)(1−q)=1+2q+2q^2 +2q^3 +.. ..+2q^( n−1) −(n−1)^2 q^( n) +n^2 q^( n+1) ⇒ (((1−q)^2 S)/q)=1+((2q(1−q^( n) ))/(1−q)) −(n−1)^2 q^( n) +n^2 q^( n+1) S = (q/((1−q)^2 ))[1+((2q(1−q^( n) ))/(1−q)) −(n−1)^2 q^n +n^2 q^(n+1) ]. (there is little error, i shall soon fix it..)

$$\frac{{S}}{{q}}=\:\mathrm{1}+\mathrm{2}^{\mathrm{2}} {q}+\mathrm{3}^{\mathrm{2}} {q}^{\mathrm{2}} +....+{n}^{\mathrm{2}} {q}^{\:{n}−\mathrm{1}} \\ $$$${S}\left(\frac{\mathrm{1}}{{q}}−\mathrm{1}\right)=\:\mathrm{1}+\mathrm{3}{q}+\mathrm{5}{q}^{\mathrm{2}} +\mathrm{7}{q}^{\mathrm{3}} +... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....+\left(\mathrm{2}{n}−\mathrm{1}\right){q}^{\:{n}−\mathrm{1}} −{n}^{\mathrm{2}} {q}^{{n}} \\ $$$${Sq}\left(\frac{\mathrm{1}}{{q}}−\mathrm{1}\right)={q}+\mathrm{3}{q}^{\mathrm{2}} +\mathrm{5}{q}^{\mathrm{3}} +... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....+\left(\mathrm{2}{n}−\mathrm{1}\right){q}^{\:{n}} −{n}^{\mathrm{2}} {q}^{{n}+\mathrm{1}} \\ $$$${S}\left(\frac{\mathrm{1}}{{q}}−\mathrm{1}\right)\left(\mathrm{1}−{q}\right)=\mathrm{1}+\mathrm{2}{q}+\mathrm{2}{q}^{\mathrm{2}} +\mathrm{2}{q}^{\mathrm{3}} +.. \\ $$$$\:\:\:\:\:\:\:..+\mathrm{2}{q}^{\:{n}−\mathrm{1}} −\left({n}−\mathrm{1}\right)^{\mathrm{2}} {q}^{\:{n}} +{n}^{\mathrm{2}} {q}^{\:{n}+\mathrm{1}} \\ $$$$\Rightarrow\:\frac{\left(\mathrm{1}−{q}\right)^{\mathrm{2}} {S}}{{q}}=\mathrm{1}+\frac{\mathrm{2}{q}\left(\mathrm{1}−{q}^{\:{n}} \right)}{\mathrm{1}−{q}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left({n}−\mathrm{1}\right)^{\mathrm{2}} {q}^{\:{n}} +{n}^{\mathrm{2}} {q}^{\:{n}+\mathrm{1}} \\ $$$$\:\:\boldsymbol{{S}}\:=\:\frac{\boldsymbol{{q}}}{\left(\mathrm{1}−\boldsymbol{{q}}\right)^{\mathrm{2}} }\left[\mathrm{1}+\frac{\mathrm{2}\boldsymbol{{q}}\left(\mathrm{1}−\boldsymbol{{q}}^{\:\boldsymbol{{n}}} \right)}{\mathrm{1}−\boldsymbol{{q}}}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left(\boldsymbol{{n}}−\mathrm{1}\right)^{\mathrm{2}} \boldsymbol{{q}}^{\boldsymbol{{n}}} +\boldsymbol{{n}}^{\mathrm{2}} \boldsymbol{{q}}^{\boldsymbol{{n}}+\mathrm{1}} \right]. \\ $$$$\:\:\:\left({there}\:{is}\:{little}\:{error},\:{i}\:{shall}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:{soon}\:{fix}\:{it}..\right) \\ $$

Commented by Saorey last updated on 22/Oct/18

thank you

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

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