$ \displaystyle a_{1}$, $ \displaystyle a_{2}$, $ \displaystyle a_{3}$ , $ \displaystyle a_{4}$ တို႔ဟာ $ \displaystyle (1 + x)^n$ ျဖန္႔လိုက္တဲ့အခါ ရလာမယ့္ ကိန္းတန္းရဲ့ အစဥ္လိုက္ရွိေသာ ကိန္း ေလးလံုး၏ (မည္သည့္ ဆက္တိုက္ ကိန္းေလးလံုးမဆို) ေျမႇာက္ေဖၚကိန္းမ်ား ျဖစ္ၾကလွ်င္
$ \displaystyle \large {\frac{{{{a}_{1}}}}{{{{a}_{1}}+{{a}_{2}}}}+\frac{{{{a}_{3}}}}{{{{a}_{3}}+{{a}_{4}}}}=\frac{{2{{a}_{2}}}}{{{{a}_{2}}+{{a}_{3}}}}}$
ျဖစ္ေၾကာင္း သက္ေသျပပါ။ မည္သည့္ကိန္းေလးလံုးမဆို လို႔ ေျပာထားပါတယ္။ မထမကိန္း ေလးလံုးလို႔ မေျပာပါဘူး။ မည့္သည့္ကိန္း ေလးလံုးမဆိုိ မွန္ကန္ရမွာ ျဖစ္ပါတယ္။ ဒါ့ေၾကာင့္တိက်တဲ့ ကိန္းေလးလံုးကို ယူၿပီး သက္ေသျပတာ ေမးခြန္းကေတာင္းဆိုခ်က္ ကို မျပည့္စံုေစပါဘူး။ မွားတယ္လို႔ ေျပာႏိုင္ပါတယ္။ general form နဲ႔ သက္ေသျပေပးရမွာ ျဖစ္ပါတယ္။
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Solution
$ \displaystyle \begin{array}{l}{{(r+1)}^{{\text{th}}}}\text{term in the expansion of}\\{{(1+x)}^{n}}={}^{n}{{C}_{r}}{{x}^{r}}\ \\\\{{a}_{1}},{{a}_{2}},{{a}_{3}}\text{ and }{{a}_{4}}\ \text{are the coefficients }\\\text{of any four consecutive terms}\text{.}\\\\\therefore {{a}_{1}}={}^{n}{{C}_{r}},{{a}_{2}}={}^{n}{{C}_{{r+1}}},\\\ \ \ {{a}_{3}}={}^{n}{{C}_{{r+2}}}\ \text{and }{{a}_{4}}={}^{n}{{C}_{{r+3}}}\end{array}$
$ \displaystyle \large{\begin{array}{l}\therefore \ \ \ \ \ \text{L}\text{.H}\text{.S}=\frac{{{{a}_{1}}}}{{{{a}_{1}}+{{a}_{2}}}}+\frac{{{{a}_{3}}}}{{{{a}_{3}}+{{a}_{4}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{}^{n}{{C}_{r}}}}{{{}^{n}{{C}_{r}}+{}^{n}{{C}_{{r+1}}}}}+\frac{{{}^{n}{{C}_{{r+2}}}}}{{{}^{n}{{C}_{{r+2}}}+{}^{n}{{C}_{{r+3}}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{}^{n}{{C}_{r}}}}{{{}^{n}{{C}_{r}}+{}^{n}{{C}_{r}}\frac{{n-(r+1)+1}}{{r+1}}}}+\frac{{{}^{n}{{C}_{{r+2}}}}}{{{}^{n}{{C}_{{r+2}}}+{}^{n}{{C}_{{r+2}}}\frac{{n-(r+3)+1}}{{r+1}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{^{n}{{C}_{r}}}}{{^{n}{{C}_{r}}\left( {1+\frac{{n-r}}{{r+1}}} \right)}}+\frac{{^{n}{{C}_{{r\text{ }+}}}_{2}}}{{^{n}{{C}_{{r\text{ }+}}}_{2}\left( {1+\frac{{n-r-2}}{{r+3}}} \right)}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{{\frac{{r+1+n-r}}{{r+1}}}}+\frac{1}{{\frac{{r+3+n-r-2}}{{r+3}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{r+1}}{{n+1}}+\frac{{r+3}}{{n+1}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2(r+2)}}{{n+1}}\\\\\therefore \ \ \ \text{R}\text{.H}\text{.S}=\frac{{2{{a}_{2}}}}{{{{a}_{2}}+{{a}_{3}}}}\ \ \ \\\ \ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2\cdot {{\ }^{n}}{{C}_{{r\text{ }+}}}_{1}}}{{^{n}{{C}_{{r\text{ }+}}}_{1}{{+}^{n}}{{C}_{{r\text{ }+}}}_{2}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2\cdot {{\ }^{n}}{{C}_{{r\text{ }+}}}_{1}}}{{^{n}{{C}_{{r\text{ }+}}}_{1}{{+}^{n}}{{C}_{{r\text{ }+}}}_{1}\frac{{n-(r+2)+1}}{{r+2}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2\cdot {{\ }^{n}}{{C}_{{r\text{ }+}}}_{1}}}{{^{n}{{C}_{{r\text{ }+}}}_{1}\left( {1+\frac{{n-r-1}}{{r+2}}} \right)}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2\cdot \ }}{{\frac{{r+2+n-r-1}}{{r+2}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2(r+2)}}{{n+1}}\\\\\therefore \frac{{{{a}_{1}}}}{{{{a}_{1}}+{{a}_{2}}}}+\frac{{{{a}_{3}}}}{{{{a}_{3}}+{{a}_{4}}}}=\frac{{2{{a}_{2}}}}{{{{a}_{2}}+{{a}_{3}}}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}}$
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