Question

If ${A_n} = \dfrac{{1.2.3 + 2.3.4 + 3.4.5 + ...{\text{ upto n terms}}}}{{n\left( {1.2 + 2.3 + 3.4 + ...{\text{ upto n terms}}} \right)}}$then $\mathop {\lim }\limits_{n \to \infty } {A_n}$ is
A) $\dfrac{3}{4}$
B) $\dfrac{1}{4}$
C) $\dfrac{1}{2}$
D) $\dfrac{5}{4}$

Answer 1

We can take the series in the numerator and the denominator. Then we can find the general term and write them under summation. Then we can expand the general terms to express them as the sum of the powers of natural numbers. Then we can apply the summation and factorize the terms. Then we can substitute the numerator and denominator and cancel the common terms. Then we can apply the limits to get the required solution.

Complete step by step solution:
We have ${A_n} = \dfrac{{1.2.3 + 2.3.4 + 3.4.5 + ...{\text{ upto n terms}}}}{{n\left( {1.2 + 2.3 + 3.4 + ...{\text{ upto n terms}}} \right)}}$.
Consider the numerator,
$N = 1.2.3 + 2.3.4 + 3.4.5 + ...{\text{ upto n terms}}$.
The 1st term is the product of the $1^{\text{st}}$ number and the numbers next and next to next to it. So, we can write the general term as,
${N_k} = k\left( {k + 1} \right)\left( {k + 2} \right)$
Now we can write the numerator as the summation of the general term from 1 to n.
$\Rightarrow N = \sum\nolimits_1^n {{N_k}}$
On substituting the value of ${N_k}$ , we get,
$\Rightarrow N = \sum\nolimits_1^n {k\left( {k + 1} \right)\left( {k + 2} \right)}$
Now we can open the brackets and do the multiplication.
$\Rightarrow N = \sum\nolimits_1^n {\left( {{k^2} + k} \right)\left( {k + 2} \right)}$
On simplifying the brackets we get,
$\Rightarrow N = \sum\nolimits_1^n {{k^3} + 2{k^2} + {k^2} + 2k}$
On Adding like terms we get,
$\Rightarrow N = \sum\nolimits_1^n {{k^3} + 3{k^2} + 2k}$
Now we can expand the summation to each term,
$\Rightarrow N = \sum\nolimits_1^n {{k^3}} + 3\sum\nolimits_1^n {{k^2}} + 2\sum\nolimits_1^n k$
We know that $\sum\nolimits_1^n {{k^3}} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$, $\sum\nolimits_1^n {{k^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$ and $\sum\nolimits_1^n k = \dfrac{{n\left( {n + 1} \right)}}{2}$ . On substituting these values, the numerator will become,
$\Rightarrow N = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + 3\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + 2\dfrac{{n\left( {n + 1} \right)}}{2}$
Now we can take the common term $\dfrac{{n\left( {n + 1} \right)}}{4}$ ,
$\Rightarrow N = \dfrac{{n\left( {n + 1} \right)}}{4}\left( {n\left( {n + 1} \right) + 2\left( {2n + 1} \right) + 4} \right)$
On simplifying the terms inside the bracket we get,
$\Rightarrow N = \dfrac{{n\left( {n + 1} \right)}}{4}\left( {{n^2} + n + 4n + 2 + 4} \right)$
On simplifying we get,
$\Rightarrow N = \dfrac{{n\left( {n + 1} \right)}}{4}\left( {{n^2} + 5n + 6} \right)$
On splitting the middle term in the bracket we get,
$\Rightarrow N = \dfrac{{n\left( {n + 1} \right)}}{4}\left( {{n^2} + 3n + 2n + 6} \right)$
On taking factors common we get,
$\Rightarrow N = \dfrac{{n\left( {n + 1} \right)}}{4}\left( {n\left( {n + 3} \right) + 2\left( {n + 3} \right)} \right)$
Hence we have,
$\Rightarrow N = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{4}$
Now we can take the denominator,
Let $D = 1.2 + 2.3 + 3.4 + ...{\text{ upto n terms}}$,
The 1st term is the product of the $1^{\text{st}}$ number and the numbers next to it. So, we can write the general term as,
${D_k} = k\left( {k + 1} \right)$
Now we can write the numerator as the summation of the general term from 1 to n.
$\Rightarrow D = \sum\nolimits_1^n {{D_k}}$
On substituting the value of ${D_k}$ , we get,
$\Rightarrow D = \sum\nolimits_1^n {k\left( {k + 1} \right)}$
Now we can open the brackets and do the multiplication.
$\Rightarrow D = \sum\nolimits_1^n {{k^2} + k}$
Now we can expand the summation to each term
$\Rightarrow D = \sum\nolimits_1^n {{k^2}} + \sum\nolimits_1^n k$
We know that $\sum\nolimits_1^n {{k^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$ and $\sum\nolimits_1^n k = \dfrac{{n\left( {n + 1} \right)}}{2}$ . On substituting these values, the denominator will become,
$\Rightarrow D = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \dfrac{{n\left( {n + 1} \right)}}{2}$
Now we can take the common factors,
$\Rightarrow D = \dfrac{{n\left( {n + 1} \right)}}{6}\left( {2n + 1 + 3} \right)$
On simplification we get,
$\Rightarrow D = \dfrac{{n\left( {n + 1} \right)}}{6}\left( {2n + 4} \right)$
Hence we have,
$\Rightarrow D = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3}$
Now we can write the given series as,
${A_n} = \dfrac{N}{{n \times D}}$
On substituting for values of N and D, we get,
${A_n} = \dfrac{{\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{4}}}{{n \times \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3}}}$
On cancelling the common terms, we get,
${A_n} = \dfrac{{3\left( {n + 3} \right)}}{{4n}}$
On simplification we get,
${A_n} = \dfrac{3}{4}\left( {1 + \dfrac{3}{n}} \right)$
We need to find $\mathop {\lim }\limits_{n \to \infty } {A_n}$ .
Let $L = \mathop {\lim }\limits_{n \to \infty } {A_n}$
$\Rightarrow L = \mathop {\lim }\limits_{n \to \infty } \dfrac{3}{4}\left( {1 + \dfrac{3}{n}} \right)$
On applying the limits, we get,
$\Rightarrow L = \dfrac{3}{4}\left( {1 + \dfrac{3}{\infty }} \right)$
We know that $\dfrac{1}{\infty } = 0$ , so the limit will become,
$\Rightarrow L = \dfrac{3}{4}\left( {1 + 0} \right)$
Hence we have,
$\Rightarrow L = \dfrac{3}{4}$
Therefore, the required limit is $\dfrac{3}{4}$.

So, the correct answer is option A.

Note:
We must simplify the numerator and denominator separately. While taking the denominator for simplification, we must not take the term n. We must make sure that this term is included while finding the limit. We cannot directly apply the limits to the given terms. While taking summations, we must take care that only addition is distributive and we cannot write the summation of the products as products of the summation.