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Question: By means of the expansion of \({e^x}\) and \({\log _e}\left( {1 + x} \right)\), when n is large, \...

By means of the expansion of ex{e^x} and log⁡e(1+x){\log _e}\left( {1 + x} \right), when n is large,
(1+1n)n=e[1−12n+abn2−716n3+....]{\left( {1 + \dfrac{1}{n}} \right)^n} = e\left[ {1 - \dfrac{1}{{2n}} + \dfrac{a}{{b{n^2}}} - \dfrac{7}{{16{n^3}}} + ....} \right], then find the value of a + b.

Explanation

Solution

Hint – In this particular question use the expansion of log (1 + x) and ex{e^x} which is given as
log⁡e(1+x)=x−x22+x33−x44+..............{\log _e}\left( {1 + x} \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + .............. and ex=1+x1!+x22!+x33!+..........{e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + .......... and later on in the solution use the property of the logarithmic i.e. log⁡eab=blog⁡ea{\log _e}{a^b} = b{\log _e}a, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
As we know that the expansion of log⁡e(1+x){\log _e}\left( {1 + x} \right) is given as
log⁡e(1+x)=x−x22+x33−x44+..............{\log _e}\left( {1 + x} \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ......................... (1)
And the expansion of ex{e^x}is
ex=1+x1!+x22!+x33!+..........{e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + .......... .............. (2)
Now consider the LHS of the given equation we have,
⇒(1+1n)n\Rightarrow {\left( {1 + \dfrac{1}{n}} \right)^n}
Let, Q=(1+1n)nQ = {\left( {1 + \dfrac{1}{n}} \right)^n}
Now take log on both sides we have,
⇒log⁡eQ=log⁡e(1+1n)n\Rightarrow {\log _e}Q = {\log _e}{\left( {1 + \dfrac{1}{n}} \right)^n}
Now use the property of logarithmic i.e. log⁡eab=blog⁡ea{\log _e}{a^b} = b{\log _e}a so use this property in the above equation we have,
⇒log⁡eQ=nlog⁡e(1+1n)\Rightarrow {\log _e}Q = n{\log _e}\left( {1 + \dfrac{1}{n}} \right)
Now from equation (1) expand log⁡e(1+1n){\log _e}\left( {1 + \dfrac{1}{n}} \right) we have,
⇒log⁡eQ=n[1n−12n2+13n3−14n4+..............]\Rightarrow {\log _e}Q = n\left[ {\dfrac{1}{n} - \dfrac{1}{{2{n^2}}} + \dfrac{1}{{3{n^3}}} - \dfrac{1}{{4{n^4}}} + ..............} \right]
Now multiply by n inside we have,
⇒log⁡eQ=[1−12n+13n2−14n3+..............]\Rightarrow {\log _e}Q = \left[ {1 - \dfrac{1}{{2n}} + \dfrac{1}{{3{n^2}}} - \dfrac{1}{{4{n^3}}} + ..............} \right]
Now take antilog on both sides we have,
⇒Q=e[1−12n+13n2−14n3+..............]\Rightarrow Q = {e^{\left[ {1 - \dfrac{1}{{2n}} + \dfrac{1}{{3{n^2}}} - \dfrac{1}{{4{n^3}}} + ..............} \right]}}
Now the above equation is also written as according to property (e1−b)=e.e−b\left( {{e^{1 - b}}} \right) = e.{e^{ - b}} so we have,
⇒Q=e.e[−12n+13n2−14n3+..............]\Rightarrow Q = e.{e^{\left[ { - \dfrac{1}{{2n}} + \dfrac{1}{{3{n^2}}} - \dfrac{1}{{4{n^3}}} + ..............} \right]}}
Now from equation (2) expand e[−12n+13n2−14n3+..............]{e^{\left[ { - \dfrac{1}{{2n}} + \dfrac{1}{{3{n^2}}} - \dfrac{1}{{4{n^3}}} + ..............} \right]}} we have,
⇒Q=e[1+[−12n+13n2−14n3+...]1!+[−12n+13n2−14n3+...]22!+[−12n+13n2−14n3+...]33!+.....]\Rightarrow Q = e\left[ {1 + \dfrac{{\left[ { - \dfrac{1}{{2n}} + \dfrac{1}{{3{n^2}}} - \dfrac{1}{{4{n^3}}} + ...} \right]}}{{1!}} + \dfrac{{{{\left[ { - \dfrac{1}{{2n}} + \dfrac{1}{{3{n^2}}} - \dfrac{1}{{4{n^3}}} + ...} \right]}^2}}}{{2!}} + \dfrac{{{{\left[ { - \dfrac{1}{{2n}} + \dfrac{1}{{3{n^2}}} - \dfrac{1}{{4{n^3}}} + ...} \right]}^3}}}{{3!}} + .....} \right]
⇒Q=e[1+(−12n+13n2−14n3+...)+12(−12n+13n2−14n3+...)2+16(−12n+13n2−14n3+...)3+.....]\Rightarrow Q = e\left[ {1 + \left( { - \dfrac{1}{{2n}} + \dfrac{1}{{3{n^2}}} - \dfrac{1}{{4{n^3}}} + ...} \right) + \dfrac{1}{2}{{\left( { - \dfrac{1}{{2n}} + \dfrac{1}{{3{n^2}}} - \dfrac{1}{{4{n^3}}} + ...} \right)}^2} + \dfrac{1}{6}{{\left( { - \dfrac{1}{{2n}} + \dfrac{1}{{3{n^2}}} - \dfrac{1}{{4{n^3}}} + ...} \right)}^3} + .....} \right]
As, 1! = 1, 2! = 2 and 3! = 6.
Now on simplifying we get,
⇒Q=e[1−12n+1124n2−716n3+..........]\Rightarrow Q = e\left[ {1 - \dfrac{1}{{2n}} + \dfrac{{11}}{{24{n^2}}} - \dfrac{7}{{16{n^3}}} + ..........} \right]
⇒(1+1n)n=e[1−12n+1124n2−716n3+..........]\Rightarrow {\left( {1 + \dfrac{1}{n}} \right)^n} = e\left[ {1 - \dfrac{1}{{2n}} + \dfrac{{11}}{{24{n^2}}} - \dfrac{7}{{16{n^3}}} + ..........} \right]
So on comparing with the given equation we have,
Therefore, (1+1n)n=e[1−12n+abn2−716n3+....]=e[1−12n+1124n2−716n3+..........]{\left( {1 + \dfrac{1}{n}} \right)^n} = e\left[ {1 - \dfrac{1}{{2n}} + \dfrac{a}{{b{n^2}}} - \dfrac{7}{{16{n^3}}} + ....} \right] = e\left[ {1 - \dfrac{1}{{2n}} + \dfrac{{11}}{{24{n^2}}} - \dfrac{7}{{16{n^3}}} + ..........} \right]
Therefore, a = 11 and b = 24.
So the value of a + b = 11 + 24 = 35.
So this is the required answer.

Note – Whenever we face such types of questions first of all using the expansion of log and exponential expand the LHS part of the given equation as above then simplify these terms as above then compare it with the RHS part of the given equation and note down the values of (a) and (b) then find out the sum of these values which is the required answer.