Answer
Verified
91.5k+ views
The sum of first n natural numbers to power one can be written as,
${{1}^{1}}+{{2}^{1}}+\ldots +{{n}^{1}}=\frac{n\left( n+1 \right)}{2}=\frac{{{n}^{2}}}{2}+\frac{n}{2}$
Similarly, the sum of first n natural numbers to power two can be written as,
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\frac{\left( {{n}^{2}}+n \right)\left( 2n+1 \right)}{6}$
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\frac{2{{n}^{3}}+{{n}^{2}}+2{{n}^{2}}+n}{6}$
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\frac{{{n}^{3}}}{3}+\frac{{{n}^{2}}}{2}+\frac{n}{6}$
The sum of first n natural numbers to power three can be written as,
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}={{\left( \frac{n\left( n+1 \right)}{2} \right)}^{2}}$
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}=\frac{{{n}^{2}}\left( {{n}^{2}}+2n+1 \right)}{4}$
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}=\frac{{{n}^{4}}+2{{n}^{3}}+{{n}^{2}}}{4}$
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}=\frac{{{n}^{4}}}{4}+\frac{{{n}^{3}}}{2}+\frac{{{n}^{2}}}{4}$
Generalizing this, we get
${{1}^{x}}+{{2}^{x}}+\ldots +{{n}^{x}}=\frac{{{n}^{x+1}}}{x+1}+{{k}_{1}}{{n}^{x}}+{{k}_{2}}{{n}^{x-1}}+\ldots $
Now substituting (x=100), we get
${{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}=\frac{{{n}^{99+1}}}{99+1}+{{k}_{1}}{{n}^{99}}+{{k}_{2}}{{n}^{99-1}}+\ldots $
Now the given expression becomes,
$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\frac{{{n}^{99+1}}}{99+1}+{{k}_{1}}{{n}^{99}}+{{k}_{2}}{{n}^{99-1}}+\ldots }{{{n}^{100}}}$
\[\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\frac{{{n}^{100}}}{100}+{{k}_{1}}{{n}^{99}}+{{k}_{2}}{{n}^{98}}+\ldots }{{{n}^{100}}}\]
\[\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{n}^{100}}}{100{{n}^{100}}}+\frac{{{k}_{1}}{{n}^{99}}}{{{n}^{100}}}+\frac{{{k}_{2}}{{n}^{98}}}{{{n}^{100}}}+\ldots \]
\[\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{100}+\frac{{{k}_{1}}}{n}+\frac{{{k}_{2}}}{{{n}^{2}}}+\ldots \]
Applying the limits, we get
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\frac{1}{100}+\frac{{{k}_{1}}}{\infty }+\frac{{{k}_{2}}}{\infty }+\ldots $
We know, $\frac{1}{\infty }\approx 0$, so
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\frac{1}{100}+0+0+\ldots $
\[\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\frac{1}{100}\]
Hence, the correct option for the given question is option (a).
Answer - Option (a)
Note - In this type of question first we have to find the summation of a given series after that check the indeterminate form of limit, then substitute the limiting value you have in your answer.
${{1}^{1}}+{{2}^{1}}+\ldots +{{n}^{1}}=\frac{n\left( n+1 \right)}{2}=\frac{{{n}^{2}}}{2}+\frac{n}{2}$
Similarly, the sum of first n natural numbers to power two can be written as,
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\frac{\left( {{n}^{2}}+n \right)\left( 2n+1 \right)}{6}$
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\frac{2{{n}^{3}}+{{n}^{2}}+2{{n}^{2}}+n}{6}$
${{1}^{2}}+{{2}^{2}}+\ldots +{{n}^{2}}=\frac{{{n}^{3}}}{3}+\frac{{{n}^{2}}}{2}+\frac{n}{6}$
The sum of first n natural numbers to power three can be written as,
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}={{\left( \frac{n\left( n+1 \right)}{2} \right)}^{2}}$
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}=\frac{{{n}^{2}}\left( {{n}^{2}}+2n+1 \right)}{4}$
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}=\frac{{{n}^{4}}+2{{n}^{3}}+{{n}^{2}}}{4}$
${{1}^{3}}+{{2}^{3}}+\ldots +{{n}^{3}}=\frac{{{n}^{4}}}{4}+\frac{{{n}^{3}}}{2}+\frac{{{n}^{2}}}{4}$
Generalizing this, we get
${{1}^{x}}+{{2}^{x}}+\ldots +{{n}^{x}}=\frac{{{n}^{x+1}}}{x+1}+{{k}_{1}}{{n}^{x}}+{{k}_{2}}{{n}^{x-1}}+\ldots $
Now substituting (x=100), we get
${{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}=\frac{{{n}^{99+1}}}{99+1}+{{k}_{1}}{{n}^{99}}+{{k}_{2}}{{n}^{99-1}}+\ldots $
Now the given expression becomes,
$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\frac{{{n}^{99+1}}}{99+1}+{{k}_{1}}{{n}^{99}}+{{k}_{2}}{{n}^{99-1}}+\ldots }{{{n}^{100}}}$
\[\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\frac{{{n}^{100}}}{100}+{{k}_{1}}{{n}^{99}}+{{k}_{2}}{{n}^{98}}+\ldots }{{{n}^{100}}}\]
\[\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{n}^{100}}}{100{{n}^{100}}}+\frac{{{k}_{1}}{{n}^{99}}}{{{n}^{100}}}+\frac{{{k}_{2}}{{n}^{98}}}{{{n}^{100}}}+\ldots \]
\[\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{100}+\frac{{{k}_{1}}}{n}+\frac{{{k}_{2}}}{{{n}^{2}}}+\ldots \]
Applying the limits, we get
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\frac{1}{100}+\frac{{{k}_{1}}}{\infty }+\frac{{{k}_{2}}}{\infty }+\ldots $
We know, $\frac{1}{\infty }\approx 0$, so
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\frac{1}{100}+0+0+\ldots $
\[\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+\ldots +{{n}^{99}}}{{{n}^{100}}}=\frac{1}{100}\]
Hence, the correct option for the given question is option (a).
Answer - Option (a)
Note - In this type of question first we have to find the summation of a given series after that check the indeterminate form of limit, then substitute the limiting value you have in your answer.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main