Question Number 56886 by gunawan last updated on 25/Mar/19
$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{r}} \frac{\mathrm{1}+{r}\:\mathrm{log}_{{e}} \mathrm{10}}{\left(\mathrm{1}+\mathrm{log}_{{e}} \mathrm{10}^{{n}} \right)^{{r}} }\:\:\mathrm{equals} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Mar/19
![a=log_e 10 and (1+nlog_e 10)=b [b=1+na] Σ_(r=0) ^n nC_r ×((1+ar)/((b)^r ))=Σ_(r=0) ^n nC_r ×(1/b^r )+aΣ_(r=0) ^n nC_r ×(r/b^r ) (1+(1/b))^n =nC_0 ×1^n +nC_1 ×(1/b^1 )+nC_2 ×(1/b^2 )+..+nC_n ×(1/b^n ) so Σ_(r=0) ^n nC_r ×(1/b^r )=(1+(1/b))^n ⇚look here aΣ_(r=0) ^n nC_r ×(r/b^r ) =a(nC_o ×(0/b^0 )+nC_1 ×(1/b^1 )+nC_2 ×(2/b^2 )+..+nC_n ×(n/b^n )) now (1+(x/b))^n =nC_0 ×(x^0 /b^0 )+nC_1 ×(x^1 /b^1 )+nC_2 ×(x^2 /b^2 )+..+nC_n ×(x^n /b^n ) differentiate both side w.r.t x n(1+(x/b))^(n−1) ×(1/b)=nC_0 ×0+nC_1 ×(1/b^1 )+nC_2 ×((2x)/b^2 )+...+nC_n ×((nx^(n−1) )/b^n ) now put x=1 both side n(1+(1/b))^(n−1) =Σ_(r=0) ^n nC_r ×(r/b^r ) so value of aΣ_(r=0) ^n nC_r ×(r/b^r )=a×n(1+(1/b))^(n−1) ⇚look here hence rewuired answer is Σ_(r=0) ^n nC_r ×(1/b^r )+aΣ_(r=0) ^n nC_n ×(r/b^r ) =(1+(1/b))^n +a×n(1+(1/b))^(n−1) =(1+(1/b))^(n−1) ×(1+(1/b)+an) [b=1+na and a=log_e 10] =(1+(1/(1+na)))^(n−1) ×(1+na+(1/(1+na))) =(1+(1/(nlog_e 10)))^(n−1) (1+nlog_e 10+(1/(1+nlog_e 10)))](https://www.tinkutara.com/question/Q56956.png)
$${a}={log}_{{e}} \mathrm{10}\:{and}\:\left(\mathrm{1}+{nlog}_{{e}} \mathrm{10}\right)={b}\:\left[{b}=\mathrm{1}+{na}\right] \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\:{nC}_{{r}} ×\frac{\mathrm{1}+{ar}}{\left({b}\right)^{{r}} }=\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{nC}_{{r}} ×\frac{\mathrm{1}}{{b}^{{r}} }+{a}\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{nC}_{{r}} ×\frac{{r}}{{b}^{{r}} } \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)^{{n}} ={nC}_{\mathrm{0}} ×\mathrm{1}^{{n}} +{nC}_{\mathrm{1}} ×\frac{\mathrm{1}}{{b}^{\mathrm{1}} }+{nC}_{\mathrm{2}} ×\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+..+{nC}_{{n}} ×\frac{\mathrm{1}}{{b}^{{n}} } \\ $$$${so}\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{nC}_{{r}} ×\frac{\mathrm{1}}{{b}^{{r}} }=\left(\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)^{{n}} \Lleftarrow{look}\:{here} \\ $$$${a}\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{nC}_{{r}} ×\frac{{r}}{{b}^{{r}} } \\ $$$$={a}\left({nC}_{{o}} ×\frac{\mathrm{0}}{{b}^{\mathrm{0}} }+{nC}_{\mathrm{1}} ×\frac{\mathrm{1}}{{b}^{\mathrm{1}} }+{nC}_{\mathrm{2}} ×\frac{\mathrm{2}}{{b}^{\mathrm{2}} }+..+{nC}_{{n}} ×\frac{{n}}{{b}^{{n}} }\right) \\ $$$${now}\:\left(\mathrm{1}+\frac{{x}}{{b}}\right)^{{n}} ={nC}_{\mathrm{0}} ×\frac{{x}^{\mathrm{0}} }{{b}^{\mathrm{0}} }+{nC}_{\mathrm{1}} ×\frac{{x}^{\mathrm{1}} }{{b}^{\mathrm{1}} }+{nC}_{\mathrm{2}} ×\frac{{x}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+..+{nC}_{{n}} ×\frac{{x}^{{n}} }{{b}^{{n}} } \\ $$$${differentiate}\:{both}\:{side}\:{w}.{r}.{t}\:{x} \\ $$$${n}\left(\mathrm{1}+\frac{{x}}{{b}}\right)^{{n}−\mathrm{1}} ×\frac{\mathrm{1}}{{b}}={nC}_{\mathrm{0}} ×\mathrm{0}+{nC}_{\mathrm{1}} ×\frac{\mathrm{1}}{{b}^{\mathrm{1}} }+{nC}_{\mathrm{2}} ×\frac{\mathrm{2}{x}}{{b}^{\mathrm{2}} }+…+{nC}_{{n}} ×\frac{{nx}^{{n}−\mathrm{1}} }{{b}^{{n}} } \\ $$$${now}\:{put}\:{x}=\mathrm{1}\:{both}\:{side} \\ $$$${n}\left(\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)^{{n}−\mathrm{1}} =\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{nC}_{{r}} ×\frac{{r}}{{b}^{{r}} } \\ $$$${so}\:{value}\:{of}\:{a}\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{nC}_{{r}} ×\frac{{r}}{{b}^{{r}} }={a}×{n}\left(\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)^{{n}−\mathrm{1}} \:\Lleftarrow{look}\:{here} \\ $$$${hence}\:{rewuired}\:{answer}\:{is} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{nC}_{{r}} ×\frac{\mathrm{1}}{{b}^{{r}} }+{a}\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{nC}_{{n}} ×\frac{{r}}{{b}^{{r}} } \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)^{{n}} +{a}×{n}\left(\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)^{{n}−\mathrm{1}} \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)^{{n}−\mathrm{1}} ×\left(\mathrm{1}+\frac{\mathrm{1}}{{b}}+{an}\right)\:\:\left[{b}=\mathrm{1}+{na}\:{and}\:{a}={log}_{{e}} \mathrm{10}\right] \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+{na}}\right)^{{n}−\mathrm{1}} ×\left(\mathrm{1}+{na}+\frac{\mathrm{1}}{\mathrm{1}+{na}}\right) \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{{nlog}_{{e}} \mathrm{10}}\right)^{{n}−\mathrm{1}} \left(\mathrm{1}+{nlog}_{{e}} \mathrm{10}+\frac{\mathrm{1}}{\mathrm{1}+{nlog}_{{e}} \mathrm{10}}\right) \\ $$