By means of the expansion of ex and log e1 - x- when n is large- 1 - 1-nn - e - 1 - 1-2n - a-bn2 -7-16n3 - - Find a-b

nbsp;Let P nbsp; =( 1 + 1n)^n nbsp; ∴ nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; log _eP = n ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Middle Terms

By means of t...

Question

By means of the expansion of $e^{x}$ and ${log}_{e} (1 + x)$ , when $n$ is large,
${(1 + \frac{1}{n})}^{n} = e [1 - \frac{1}{2 n} + \frac{a}{b n^{2}} - \frac{7}{16 n^{3}} + . . .]$
Find $a + b$

Open in App

Solution

Let P $= {(1 + \frac{1}{n})}^{n}$
$∴$ ${log}_{e} P = n {log}_{e} (1 + \frac{1}{n})$
$= n (\frac{1}{n} - \frac{1}{2 n^{2}} + \frac{1}{3 n^{3}} - \frac{1}{4 n^{4}} + . . .)$
$= (1 - \frac{1}{2 n} + \frac{1}{3 n^{2}} - \frac{1}{4 n^{3}} + . . .)$
$∴ P = e^{(1 - \frac{1}{2 n} + \frac{1}{3 n^{2}} - \frac{1}{4 n^{3}} + . . .)}$
$= e . e^{⎛ ⎝ - \frac{1}{2 n} + \frac{1}{3 n^{2}} - \frac{1}{4 n^{3}} + . . . \infty ⎞ ⎠}$
$= e [1 - (\frac{1}{2 n} - \frac{1}{3 n^{2}} + \frac{1}{4 n^{4}} - . . .) + \frac{1}{2!} {(\frac{1}{2 n} - \frac{1}{3 n^{2}} + \frac{1}{4 n^{4}} - . . .)}^{2} - \frac{1}{3!} {(\frac{1}{2 n} - \frac{1}{3 n^{2}} + \frac{1}{4 n^{4}} - . . .)}^{3}]$
$= e [1 - \frac{1}{2 n} + \frac{11}{24 n^{2}} - \frac{7}{16 n^{3}} + . . .]$
Hence
${(1 + \frac{1}{n})}^{n} = e [1 - \frac{1}{2 n} + \frac{11}{24 n^{2}} - \frac{7}{16 n^{3}} + . . .]$
$∴ a + b = 35$

Suggest Corrections

Similar questions

Q. If each coefficient in the expansion of the expression

x (1 + x)^{n} (n ϵ N)

in powers of 'x' is divided by the exponent of corresponding power, then the sum of the values thus obtained is equal to

If $x < 1$ , then the coefficient of $x^{n}$ in the expansion of $\frac{e^{x}}{1 - x}$ is $1 + \frac{a}{1!} + \frac{a^{2}}{2!} + . . . + \frac{a^{n - 2}}{(n - 2)!} + \frac{a^{n - 1}}{(n - 1)!}$ .

Find $a + 7$

Q. The coefficient of

\frac{1}{x}

in the expansion of

(1 + x)^{n} {(1 + \frac{1}{x})}^{n}

Q. The coefficient of

x^{r} (0 \leq r \leq n - 1)

in the expansion of

E = {(x + 2)}^{n - 1} + {(x + 2)}^{n - 2} (x + 1) + {(x + 2)}^{n - 3} {(x + 1)}^{2} + . . . . + {(x + 1)}^{n - 1}

$n^{n} - n (n - 1)^{n} + \frac{n (n - 1)}{2!} (n - 2)^{n} - . . . \infty = \frac{3}{a} \times n!$ .

Find $a$

Join BYJU'S Learning Program

By means of the expansion of ex and loge(1+x), when n is large,(1+1n)n=e[1−12n+abn2−716n3+...] Find a+b

By means of the expansion of $e^{x}$ and ${log}_{e} (1 + x)$ , when $n$ is large,
${(1 + \frac{1}{n})}^{n} = e [1 - \frac{1}{2 n} + \frac{a}{b n^{2}} - \frac{7}{16 n^{3}} + . . .]$
Find $a + b$