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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

$${}_1{F}_a(1;\frac{a+b}{a},\frac{a+b-1}{a},.......,\frac{b+1}{a};\frac{x}{a^a})$$$$=\sum _{k⩾0}\frac{k!}{\underset{j=0}{\prod ^{a-1}}\underset{l=0}{\prod ^{k-1}}(\frac{a+b-j}{a}+l)}.\frac{x^k}{a^{ak}}.\frac{1}{k!}$$$$=\sum _{k⩾0}\frac{x^k}{\underset{j=0}{\prod ^{a-1}}\underset{l=0}{\prod ^{k-1}}(a+b-j+al).\frac{1}{a^{k.a}}}.\frac{1}{a^{ka}}$$$$=\sum _{k⩾0}\frac{x^k}{\underset{j=0}{\prod ^{a-1}}\underset{l=0}{\prod ^{k-1}}(a+b-j+al)}=\sum _{k⩾0}\frac{x^k}{\underset{l=0}{\prod ^{k-1}}\underset{j=0}{\prod ^{a-1}}(a(1+l)+b-j)}$$$$=\sum _{k⩾0}\frac{x^k}{\underset{l=0}{\prod ^{k-1}}\frac{(a(1+l)+b)!}{(al+b)!}}$$$$=\sum _{k⩾0}\frac{x^k}{\underset{l=0}{\prod ^{k-1}}(a(1+l)+b)!.\frac{1}{\left(b\right)!.\underset{l=0}{\prod ^{k-2}}(a(l+1)+b)!}}$$$$\sum _{k⩾0}\frac{b!x^k}{(ak+b)!.}=b!.\sum _{k⩾0}\frac{x^k}{(ak+b)!}therisb!$$

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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