Question-81222

Question Number 81222 by M±th+et£s last updated on 10/Feb/20

Answered by mind is power last updated on 10/Feb/20

_1 F_a (1;((a+b)/a),((a+b−1)/a),.......,((b+1)/a);(x/a^a )) =Σ_(k≥0) ((k!)/(Π_(j=0) ^(a−1) Π_(l=0) ^(k−1) (((a+b−j)/a)+l))).(x^k /a^(ak) ).(1/(k!)) =Σ_(k≥0) (x^k /(Π_(j=0) ^(a−1) Π_(l=0) ^(k−1) (a+b−j+al).(1/a^(k.a) ))).(1/a^(ka) ) =Σ_(k≥0) (x^k /(Π_(j=0) ^(a−1) Π_(l=0) ^(k−1) (a+b−j+al)))=Σ_(k≥0) (x^k /(Π_(l=0) ^(k−1) Π_(j=0) ^(a−1) (a(1+l)+b−j))) =Σ_(k≥0) (x^k /(Π_(l=0) ^(k−1) (((a(1+l)+b)!)/((al+b)!)))) =Σ_(k≥0) (x^k /(Π_(l=0) ^(k−1) (a(1+l)+b)!.(1/((b)!.Π_(l=0) ^(k−2) (a(l+1)+b)!)))) Σ_(k≥0) ((b!x^k )/((ak+b)!.))=b!.Σ_(k≥0) (x^k /((ak+b)!)) ther is b!

$$\:\:_{\mathrm{1}} {F}_{{a}} \left(\mathrm{1};\frac{{a}+{b}}{{a}},\frac{{a}+{b}−\mathrm{1}}{{a}},…….,\frac{{b}+\mathrm{1}}{{a}};\frac{{x}}{{a}^{{a}} }\right) \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{k}!}{\underset{{j}=\mathrm{0}} {\overset{{a}−\mathrm{1}} {\prod}}\underset{{l}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left(\frac{{a}+{b}−{j}}{{a}}+{l}\right)}.\frac{{x}^{{k}} }{{a}^{{ak}} }.\frac{\mathrm{1}}{{k}!} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{x}^{{k}} }{\underset{{j}=\mathrm{0}} {\overset{{a}−\mathrm{1}} {\prod}}\underset{{l}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left({a}+{b}−{j}+{al}\right).\frac{\mathrm{1}}{{a}^{{k}.{a}} }}.\frac{\mathrm{1}}{{a}^{{ka}} } \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{x}^{{k}} }{\underset{{j}=\mathrm{0}} {\overset{{a}−\mathrm{1}} {\prod}}\underset{{l}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left({a}+{b}−{j}+{al}\right)}=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{x}^{{k}} }{\underset{{l}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\underset{{j}=\mathrm{0}} {\overset{{a}−\mathrm{1}} {\prod}}\left({a}\left(\mathrm{1}+{l}\right)+{b}−{j}\right)} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{x}^{{k}} }{\underset{{l}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\frac{\left({a}\left(\mathrm{1}+{l}\right)+{b}\right)!}{\left({al}+{b}\right)!}} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{x}^{{k}} }{\underset{{l}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left({a}\left(\mathrm{1}+{l}\right)+{b}\right)!.\frac{\mathrm{1}}{\left({b}\right)!.\underset{{l}=\mathrm{0}} {\overset{{k}−\mathrm{2}} {\prod}}\left({a}\left({l}+\mathrm{1}\right)+{b}\right)!}} \\ $$$$\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{b}!{x}^{{k}} }{\left({ak}+{b}\right)!.}={b}!.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{x}^{{k}} }{\left({ak}+{b}\right)!}\:\:{ther}\:{is}\:{b}!\:\: \\ $$

Commented by M±th+et£s last updated on 11/Feb/20

$$\mathrm{you}\:\mathrm{are}\:\mathrm{right}\:\mathrm{sir}\:\mathrm{its}\:\mathrm{my}\:\mathrm{fault} \\ $$

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