Question Number 60745 by Tawa1 last updated on 25/May/19
Commented by Prithwish sen last updated on 25/May/19
$$\frac{\left[\left(\mathrm{n}+\mathrm{1}\right)!\right]^{\mathrm{n}} }{\left(\mathrm{n}!\right)^{\mathrm{n}+\mathrm{1}} }\:=\left[\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\mathrm{n}!}\right]^{\mathrm{n}} .\frac{\mathrm{1}}{\mathrm{n}!}\:=\frac{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}!} \\ $$$$=\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{1}}.\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}…………\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{n}}\:>\mathrm{1} \\ $$$$\therefore\:\left[\left(\mathrm{n}+\mathrm{1}\right)!\right]^{\mathrm{n}} \:>\left(\mathrm{n}!\right)^{\mathrm{n}+\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{i}.\mathrm{e}\left(\:\mathrm{101}!\right)^{\mathrm{100}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{larger} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 25/May/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by maxmathsup by imad last updated on 25/May/19
$$\left.\mathrm{3}\right)\:{we}\:{have}\:\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{100}} \:=\sum_{{k}=\mathrm{0}} ^{\mathrm{100}} \:{C}_{\mathrm{100}} ^{{k}} \:\left(\mathrm{2}{x}\right)^{{k}} \left(−\mathrm{1}\right)^{\mathrm{100}−{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{100}} \:{C}_{\mathrm{100}} ^{{k}} \:\mathrm{2}^{{k}} \left(−\mathrm{1}\right)^{{k}} \:{x}^{{k}} \:=\sum_{{k}=\mathrm{0}} ^{\mathrm{100}} \:{a}_{{k}} \:{x}^{{k}} \:\:\Rightarrow{a}_{{k}} =\:{C}_{\mathrm{100}} ^{{k}} \:\left(−\mathrm{2}\right)^{{k}} \:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{\mathrm{100}} \:\:{a}_{{k}} =\sum_{{k}=\mathrm{0}} ^{\mathrm{100}} \:{C}_{\mathrm{100}} ^{{k}} \left(−\mathrm{2}\right)^{{k}} \:\left(\mathrm{1}\right)^{\mathrm{100}−{k}} \:=\left(−\mathrm{2}+\mathrm{1}\right)^{\mathrm{100}} \:=\mathrm{1}\:. \\ $$
Commented by Tawa1 last updated on 25/May/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by maxmathsup by imad last updated on 25/May/19
$${you}\:{are}\:{welcome}\:{sir}\:. \\ $$
Answered by tanmay last updated on 25/May/19
$$\left.\mathrm{2}\right)\left(\mathrm{2}{y}−{x}\right)^{{k}} =\left(\mathrm{2}{y}\right)^{{k}} −{kc}_{\mathrm{1}} \left(\mathrm{2}{y}\right)^{{k}−\mathrm{1}} \left({x}\right)+{kc}_{\mathrm{2}} \left(\mathrm{2}{y}\right)^{{k}−\mathrm{2}} \left({x}\right)^{\mathrm{2}} −{kc}_{\mathrm{3}} \left(\mathrm{2}{y}\right)^{{k}−\mathrm{3}} {x}^{\mathrm{3}} +…\left(−\mathrm{1}\right)^{{k}} \left({x}\right)^{{k}} \\ $$$${so}\:{sum}\:{of}\:{coefficients} \\ $$$$=\mathrm{2}^{{k}} −{kc}_{\mathrm{1}} \left(\mathrm{2}\right)^{{k}−\mathrm{1}} +{kc}_{\mathrm{2}} \mathrm{2}^{{k}−\mathrm{2}} −{kc}_{\mathrm{3}} \mathrm{2}^{{k}−\mathrm{3}} +..+\left(−\mathrm{1}\right)^{{k}} \\ $$$$=\left(\mathrm{2}×\mathrm{1}−\mathrm{1}\right)^{{k}} \:\:\:\left[{put}\:{x}=\mathrm{1}\:\:{and}\:\:{y}=\mathrm{1}\right] \\ $$$$=\mathrm{1}\:\:\: \\ $$$$ \\ $$
Commented by Tawa1 last updated on 25/May/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by tanmay last updated on 25/May/19
$$\left.\mathrm{3}\right)\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{100}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+…+{a}_{\mathrm{100}} {x}^{\mathrm{100}} \\ $$$${now}\:{put}\:{x}=\mathrm{1} \\ $$$$\left(\mathrm{1}\right)^{\mathrm{100}} =\mathrm{1}={sum}\:{of}\:{coefficient} \\ $$
Commented by Tawa1 last updated on 25/May/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by tanmay last updated on 25/May/19
$$\left.\mathrm{3}\right) \\ $$$$\left.{ii}\right)\left({x}^{\mathrm{3}} −{x}−\mathrm{1}\right)^{\mathrm{99}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +…+{a}_{{i}} {x}^{\mathrm{297}} \\ $$$${put}\:{x}=\mathrm{1}\: \\ $$$${sum}\:{of}\:{coefficient}=\left(\mathrm{1}^{\mathrm{3}} −\mathrm{1}−\mathrm{1}\right)^{\mathrm{99}} =−\mathrm{1} \\ $$
Commented by Tawa1 last updated on 25/May/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by tanmay last updated on 25/May/19
$$\left.\mathrm{4}\right) \\ $$$$\left(\mathrm{2}{x}+\mathrm{3}{y}−{z}\right)^{\mathrm{20}} \\ $$$${put}\:{y}=\mathrm{1}\:{z}=\mathrm{2} \\ $$$$\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{20}} ={a}_{\mathrm{0}} {x}^{\mathrm{20}} +{a}_{\mathrm{1}} {x}^{\mathrm{19}} +{a}_{\mathrm{2}} {x}^{\mathrm{18}} +…+\mathrm{1} \\ $$$${now}\:{put}\:{y}=−\mathrm{1}\:\:{z}=−\mathrm{2} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{20}} ={a}_{\mathrm{0}} {x}^{\mathrm{20}} −{a}_{\mathrm{1}} {x}^{\mathrm{19}} +{a}_{\mathrm{2}} {x}^{\mathrm{18}} +…+\mathrm{1} \\ $$$$\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{20}} −\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{20}} \\ $$$$=\mathrm{2}\left[{a}_{\mathrm{1}} {x}^{\mathrm{19}} +{a}_{\mathrm{3}} {x}^{\mathrm{17}} +…\right] \\ $$$${put}\:{x}=\mathrm{1}\:{both}\:{side} \\ $$$${so}\:{sum}\:{of}\:{coefficient}\:{containing}\:{odd}\:{power}\:{of} \\ $$$${x}\:{is} \\ $$$$=\frac{\mathrm{3}^{\mathrm{20}} −\mathrm{1}^{\mathrm{20}} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}^{\mathrm{20}} −\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Prithwish sen last updated on 25/May/19
$$\mathrm{excellent}\:\mathrm{sir}\:. \\ $$
Commented by tanmay last updated on 25/May/19
$${thank}\:{you}\:{sir}… \\ $$
Commented by Tawa1 last updated on 25/May/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 25/May/19
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir} \\ $$