Answer

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Hint: Try expanding the series and then observing the terms individually.

The above given term is

$\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{x=1}^{20}{{{\cos }^{2n}}(x-10)}$

Take the part that comes after the summation sign, or the series whose general term is given, and open the series. Doing so, we get :

$=\underset{n\to \infty }{\mathop{\lim }}\,[{{\cos }^{2n}}(-9)+{{\cos }^{2n}}(-8)+{{\cos }^{2n}}(-7)+........+{{\cos }^{2n}}(10)]$

Now, let’s see the possible values $x-10$ can have according to the lower and upper limit of the summation given. The lower limit given to us is $x=1$. Which means, that the lowest value of $x-10$ that we can have is $1-10=-9$, and since the upper limit given to us is $x=20$, the highest value of the term $x-10$ will be equal to $20-10=10$. Thus, we can notice, that as we increase the value of $x$ being substituted, the term gets bigger in value, i.e. all the terms in the series are basically ${{\cos }^{2n}}(-9),{{\cos }^{2n}}(-8),.........,{{\cos }^{2n}}(9),{{\cos }^{2n}}(10)$.

As, from these series one value that comes in between will definitely be $={{\cos }^{2n}}(0)........(i)$

The range of cos functions is $\left( -1<\cos x\le 1 \right)$.

This means that the cosine of every argument passed to the function will have its absolute value between $0$ and $1$ only.

Also, we know if increase the power until $\infty $, for any value between $0$ and $1$, the value becomes $0$, Thus, for all the terms except ${{\cos }^{2n}}(0)$, its value on putting the limit will be $0$.

Therefore, ultimately, $\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{x=1}^{20}{{{\cos }^{2n}}(x-10)}$

$=0+0+0+.......+{{\cos }^{2n}}(0)+....+0+0.$

Hence, except ${{\cos }^{2n}}(0)$ all other values there are 0.

Now, come to the term ${{\cos }^{2n}}(0)$ and check the form.

$\underset{n\to \infty }{\mathop{\lim }}\,{{\left[ \cos \left( x-10 \right) \right]}^{2n}}$

Substituting $\left( x=10 \right)$, we get; ${{\left( \cos 0 \right)}^{2\infty }}\Rightarrow {{\left( 1 \right)}^{\infty }}$. As, $\left( {{1}^{\infty }}=0 \right)$, (by taking the value little <1)

Hence, 0 is the correct answer of the $\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{x=1}^{20}{{{\cos }^{2n}}(x-10)}$

Note: Don’t get confused between actual and absolute values. Due to the range of the cosine function, we could say that the range of their absolute values always lies between $0$ and $1$.

The above given term is

$\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{x=1}^{20}{{{\cos }^{2n}}(x-10)}$

Take the part that comes after the summation sign, or the series whose general term is given, and open the series. Doing so, we get :

$=\underset{n\to \infty }{\mathop{\lim }}\,[{{\cos }^{2n}}(-9)+{{\cos }^{2n}}(-8)+{{\cos }^{2n}}(-7)+........+{{\cos }^{2n}}(10)]$

Now, let’s see the possible values $x-10$ can have according to the lower and upper limit of the summation given. The lower limit given to us is $x=1$. Which means, that the lowest value of $x-10$ that we can have is $1-10=-9$, and since the upper limit given to us is $x=20$, the highest value of the term $x-10$ will be equal to $20-10=10$. Thus, we can notice, that as we increase the value of $x$ being substituted, the term gets bigger in value, i.e. all the terms in the series are basically ${{\cos }^{2n}}(-9),{{\cos }^{2n}}(-8),.........,{{\cos }^{2n}}(9),{{\cos }^{2n}}(10)$.

As, from these series one value that comes in between will definitely be $={{\cos }^{2n}}(0)........(i)$

The range of cos functions is $\left( -1<\cos x\le 1 \right)$.

This means that the cosine of every argument passed to the function will have its absolute value between $0$ and $1$ only.

Also, we know if increase the power until $\infty $, for any value between $0$ and $1$, the value becomes $0$, Thus, for all the terms except ${{\cos }^{2n}}(0)$, its value on putting the limit will be $0$.

Therefore, ultimately, $\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{x=1}^{20}{{{\cos }^{2n}}(x-10)}$

$=0+0+0+.......+{{\cos }^{2n}}(0)+....+0+0.$

Hence, except ${{\cos }^{2n}}(0)$ all other values there are 0.

Now, come to the term ${{\cos }^{2n}}(0)$ and check the form.

$\underset{n\to \infty }{\mathop{\lim }}\,{{\left[ \cos \left( x-10 \right) \right]}^{2n}}$

Substituting $\left( x=10 \right)$, we get; ${{\left( \cos 0 \right)}^{2\infty }}\Rightarrow {{\left( 1 \right)}^{\infty }}$. As, $\left( {{1}^{\infty }}=0 \right)$, (by taking the value little <1)

Hence, 0 is the correct answer of the $\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{x=1}^{20}{{{\cos }^{2n}}(x-10)}$

Note: Don’t get confused between actual and absolute values. Due to the range of the cosine function, we could say that the range of their absolute values always lies between $0$ and $1$.

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