Question

The value of \[\dfrac{2}{1!}+\dfrac{2+4}{2!}+\dfrac{2+4+6}{3!}+.......\infty \] is
(a) e
(b) 2e
(c) 3e
(d) none of these

Answer 1

Hint: In this question, let us first write the general form of any term in the given sequence. Then, the numerator in the given sequence is in A.P which can be simplified using the formula \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]. Now, cancel the common terms of the numerator and denominator and then express the series in a simplified form . Now, use the expansion \[e=1+\dfrac{1}{1!}+\dfrac{1}{2!}+.....\] and convert accordingly to get the result.


Complete step-by-step solution -
Now, the given series from the question is
\[\dfrac{2}{1!}+\dfrac{2+4}{2!}+\dfrac{2+4+6}{3!}+.......\infty \]
Let us now write the general form of any \[{{n}^{th}}\]term of this series
\[{{T}_{n}}=\dfrac{2+4+6+...+2n}{n!}\]
Now, we can observe that the numerator in the above expression seems to be in A.P
As we already know that from the formula of sum of n terms of an A.P that
\[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]
Now, from the above formula on comparison we have
\[a=2,d=2\]
Now, on substituting the respective values in the \[{{n}^{th}}\]term of this series we get,
\[{{T}_{n}}=\dfrac{\dfrac{n}{2}\left[ 2\times 2+\left( n-1 \right)2 \right]}{n!}\]
Now, on further simplification we get,
\[{{T}_{n}}=\dfrac{\dfrac{n}{2}\left[ 4+2n-2 \right]}{n!}\]
Now, on cancelling the common terms and simplifying further we get,
\[{{T}_{n}}=\dfrac{n\left( n+1 \right)}{n!}\]
Now, this can be further written as
\[{{T}_{n}}=\dfrac{n}{\left( n-1 \right)!}+\dfrac{1}{\left( n-1 \right)!}\]
Now, the given series can be written as
\[\Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{n}{\left( n-1 \right)!}+\dfrac{1}{\left( n-1 \right)!}}\]
Let us now further write it in the simplified form
\[\Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{n-1+1}{\left( n-1 \right)!}}+\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-1 \right)!}}\]
Now, on further simplification we can write it as
\[\Rightarrow \sum\limits_{n=2}^{\infty }{\dfrac{1}{\left( n-2 \right)!}}+\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-1 \right)!}}+\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-1 \right)!}}\]
Now, this can be further written as
\[\Rightarrow \sum\limits_{n=2}^{\infty }{\dfrac{1}{\left( n-2 \right)!}}+2\sum\limits_{n=1}^{\infty }{\dfrac{1}{\left( n-1 \right)!}}\]
Now, let us expand each of the terms by expanding them respectively and simplify further
\[\Rightarrow 1+\dfrac{1}{1!}+\dfrac{1}{2!}+.....+2\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+..... \right)\]
Now, this can be also written as
\[\Rightarrow 3\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+..... \right)\]
Here, as we already know that
\[e=1+\dfrac{1}{1!}+\dfrac{1}{2!}+.....\]
Now, on substituting the respective value in the above expression we get,
\[\Rightarrow 3e\]
Hence, the correct option is (c).

Note: Instead of writing in terms of the sum of n terms of an A.P we can also solve it but finding general terms and then simplify further. But, it would take a lot of time. So, using the series formula would be better. It is important to note that we need to convert the numerator of the general to 1 so that we can express it in terms of e expansion. While simplifying we should not neglect any of the terms or forget to convert the numerator to 1 because it will be difficult to convert into e.