r-0-n-n-C-r-1-r-log-e-10-1-log-e-10-n-r-equals-

Question Number 56886 by gunawan last updated on 25/Mar/19

\underset{r = 0}{\sum^{n}}^{n} C_{r} \frac{1 + r \log_{e} 10}{{(1 + \log_{e} 10^{n})}^{r}} equals

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Mar/19

a = {l o g}_{e} 10 a n d (1 + {n l o g}_{e} 10) = b [b = 1 + n a]

\underset{r = 0}{\sum^{n}} {n C}_{r} \times \frac{1 + a r}{{(b)}^{r}} = \underset{r = 0}{\sum^{n}} {n C}_{r} \times \frac{1}{b^{r}} + a \underset{r = 0}{\sum^{n}} {n C}_{r} \times \frac{r}{b^{r}}

{(1 + \frac{1}{b})}^{n} = {n C}_{0} \times 1^{n} + {n C}_{1} \times \frac{1}{b^{1}} + {n C}_{2} \times \frac{1}{b^{2}} + . . + {n C}_{n} \times \frac{1}{b^{n}}

s o \underset{r = 0}{\sum^{n}} {n C}_{r} \times \frac{1}{b^{r}} = {(1 + \frac{1}{b})}^{n} ⇚ l o o k h e r e

a \underset{r = 0}{\sum^{n}} {n C}_{r} \times \frac{r}{b^{r}}

= a ({n C}_{o} \times \frac{0}{b^{0}} + {n C}_{1} \times \frac{1}{b^{1}} + {n C}_{2} \times \frac{2}{b^{2}} + . . + {n C}_{n} \times \frac{n}{b^{n}})

n o w {(1 + \frac{x}{b})}^{n} = {n C}_{0} \times \frac{x^{0}}{b^{0}} + {n C}_{1} \times \frac{x^{1}}{b^{1}} + {n C}_{2} \times \frac{x^{2}}{b^{2}} + . . + {n C}_{n} \times \frac{x^{n}}{b^{n}}

d i f f e r e n t i a t e b o t h s i d e w . r . t x

n {(1 + \frac{x}{b})}^{n - 1} \times \frac{1}{b} = {n C}_{0} \times 0 + {n C}_{1} \times \frac{1}{b^{1}} + {n C}_{2} \times \frac{2 x}{b^{2}} + \dots + {n C}_{n} \times \frac{{n x}^{n - 1}}{b^{n}}

n o w p u t x = 1 b o t h s i d e

n {(1 + \frac{1}{b})}^{n - 1} = \underset{r = 0}{\sum^{n}} {n C}_{r} \times \frac{r}{b^{r}}

s o v a l u e o f a \underset{r = 0}{\sum^{n}} {n C}_{r} \times \frac{r}{b^{r}} = a \times n {(1 + \frac{1}{b})}^{n - 1} ⇚ l o o k h e r e

h e n c e r e w u i r e d a n s w e r i s

\underset{r = 0}{\sum^{n}} {n C}_{r} \times \frac{1}{b^{r}} + a \underset{r = 0}{\sum^{n}} {n C}_{n} \times \frac{r}{b^{r}}

= {(1 + \frac{1}{b})}^{n} + a \times n {(1 + \frac{1}{b})}^{n - 1}

= {(1 + \frac{1}{b})}^{n - 1} \times (1 + \frac{1}{b} + a n) [b = 1 + n a a n d a = {l o g}_{e} 10]

= {(1 + \frac{1}{1 + n a})}^{n - 1} \times (1 + n a + \frac{1}{1 + n a})

= {(1 + \frac{1}{{n l o g}_{e} 10})}^{n - 1} (1 + {n l o g}_{e} 10 + \frac{1}{1 + {n l o g}_{e} 10})