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Question Number 127508 by mohammad17 last updated on 30/Dec/20

Commented by mohammad17 last updated on 30/Dec/20

$$suchthat\beta betafunction$$

Answered by Ar Brandon last updated on 30/Dec/20

$$\beta (\mathrm{a},\mathrm{b})\beta (\mathrm{a}+\mathrm{b},\mathrm{c})$$$$=\frac{\mathrm{\Gamma }\left(\mathrm{a}\right)\mathrm{\Gamma }\left(\mathrm{b}\right)}{\mathrm{\Gamma }(\mathrm{a}+\mathrm{b})}\cdot \frac{\mathrm{\Gamma }(\mathrm{a}+\mathrm{b})\mathrm{\Gamma }\left(\mathrm{c}\right)}{\mathrm{\Gamma }(\mathrm{a}+\mathrm{b}+\mathrm{c})}$$$$=\frac{\mathrm{\Gamma }\left(\mathrm{a}\right)\mathrm{\Gamma }\left(\mathrm{b}\right)\mathrm{\Gamma }\left(\mathrm{c}\right)}{\mathrm{\Gamma }(\mathrm{a}+\mathrm{b}+\mathrm{c})}$$$$=\frac{\mathrm{\Gamma }\left(\mathrm{a}\right)\mathrm{\Gamma }(\mathrm{b}+\mathrm{c})}{\mathrm{\Gamma }(\mathrm{a}+\mathrm{b}+\mathrm{c})}\cdot \frac{\mathrm{\Gamma }\left(\mathrm{b}\right)\mathrm{\Gamma }\left(\mathrm{c}\right)}{\mathrm{\Gamma }(\mathrm{b}+\mathrm{c})}$$$$=\beta (\mathrm{a},\mathrm{b}+\mathrm{c})\beta (\mathrm{b},\mathrm{c})$$

Commented by mohammad17 last updated on 30/Dec/20

$$thankyousircanyouhelpmeinallquestion$$

Answered by Ar Brandon last updated on 30/Dec/20

$$\beta (\mathrm{a},\mathrm{b})\beta (\mathrm{a}+\mathrm{b},\mathrm{c})$$$$=\frac{\mathrm{\Gamma }\left(\mathrm{a}\right)\mathrm{\Gamma }\left(\mathrm{b}\right)}{\mathrm{\Gamma }(\mathrm{a}+\mathrm{b})}\cdot \frac{\mathrm{\Gamma }(\mathrm{a}+\mathrm{b})\mathrm{\Gamma }\left(\mathrm{c}\right)}{\mathrm{\Gamma }(\mathrm{a}+\mathrm{b}+\mathrm{c})}$$$$=\frac{\mathrm{\Gamma }\left(\mathrm{a}\right)\mathrm{\Gamma }\left(\mathrm{b}\right)\mathrm{\Gamma }\left(\mathrm{c}\right)}{\mathrm{\Gamma }(\mathrm{a}+\mathrm{b}+\mathrm{c})}$$$$=\frac{\mathrm{\Gamma }\left(\mathrm{b}\right)\mathrm{\Gamma }(\mathrm{a}+\mathrm{c})}{\mathrm{\Gamma }(\mathrm{a}+\mathrm{b}+\mathrm{c})}\cdot \frac{\mathrm{\Gamma }\left(\mathrm{a}\right)\mathrm{\Gamma }\left(\mathrm{c}\right)}{\mathrm{\Gamma }(\mathrm{a}+\mathrm{c})}$$$$=\beta (\mathrm{a},\mathrm{c})\beta (\mathrm{b},\mathrm{c}+\mathrm{a})$$

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