Problem Study : Binomial Theorem (ေမးခြန္းေဟာင္း)

 If $\displaystyle a_{1}$, $\displaystyle a_{2}$, $\displaystyle a_{3}$ and $\displaystyle a_{4}$ are the coefficients of any four consecutive terms in the expansion of $\displaystyle (1 + x)^n$, show that $\Large {\frac{{{{a}_{1}}}}{{{{a}_{1}}+{{a}_{2}}}}+\frac{{{{a}_{3}}}}{{{{a}_{3}}+{{a}_{4}}}}=\frac{{2{{a}_{2}}}}{{{{a}_{2}}+{{a}_{3}}}}}$.

$\displaystyle a_{1}$, $\displaystyle a_{2}$, $\displaystyle a_{3}$ , $\displaystyle a_{4}$ 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 မ​ေျပာပါဘူး။ မည့္သည့္ကိန္း ေလးလံုးမဆိုိ မွန္ကန္ရမွာ ျဖစ္ပါတယ္။ ဒါ့ေၾကာင့္တိက်တဲ့ ကိန္းေလးလံုးကို ယူၿပီး သက္ေသျပတာ ေမးခြန္းကေတာင္းဆိုခ်က္ ကို မျပည့္စံုေစပါဘူး။ မွားတယ္လို႔ ေျပာႏိုင္ပါတယ္။ general form နဲ႔ သက္ေသျပေပးရမွာ ျဖစ္ပါတယ္။

------------------------------------------------------------------------ 

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{$$\begin{array}{l}\therefore \text{L}\text{.H}\text{.S}=\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}\\ \\ =\frac{{}^n{C}_r}{{}^n{C}_r+{}^n{C}_{r+1}}+\frac{{}^n{C}_{r+2}}{{}^n{C}_{r+2}+{}^n{C}_{r+3}}\\ \\ =\frac{{}^n{C}_r}{{}^n{C}_r+{}^n{C}_r\frac{n-(r+1)+1}{r+1}}+\frac{{}^n{C}_{r+2}}{{}^n{C}_{r+2}+{}^n{C}_{r+2}\frac{n-(r+3)+1}{r+1}}\\ \\ =\frac{{}^n{C}_r}{{}^n{C}_r\left(1+\frac{n-r}{r+1}\right)}+\frac{{}^n{C_{r+}}_2}{{}^n{C_{r+}}_2\left(1+\frac{n-r-2}{r+3}\right)}\\ \\ =\frac{1}{\frac{r+1+n-r}{r+1}}+\frac{1}{\frac{r+3+n-r-2}{r+3}}\\ \\ =\frac{r+1}{n+1}+\frac{r+3}{n+1}\\ \\ =\frac{2(r+2)}{n+1}\\ \\ \therefore \text{R}\text{.H}\text{.S}=\frac{2a_2}{a_2+a_3}\\ \\ =\frac{2\cdot {}^n{C_{r+}}_1}{{}^n{C_{r+}}_1{+}^n{C_{r+}}_2}\\ \\ =\frac{2\cdot {}^n{C_{r+}}_1}{{}^n{C_{r+}}_1{+}^n{C_{r+}}_1\frac{n-(r+2)+1}{r+2}}\\ \\ =\frac{2\cdot {}^n{C_{r+}}_1}{{}^n{C_{r+}}_1\left(1+\frac{n-r-1}{r+2}\right)}\\ \\ =\frac{2\cdot }{\frac{r+2+n-r-1}{r+2}}\\ \\ =\frac{2(r+2)}{n+1}\\ \\ \therefore \frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{2a_2}{a_2+a_3}\\ \end{array}$$}$

စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!