Question

If $1.\left( 0! \right)+3.\left( 1! \right)+7.\left( 2! \right)+13.\left( 3! \right)+21.\left( 4! \right)......$ up to n terms = $\left( 4000 \right)4000!$ then the value of n is?
(a) 4000
(b) 4001
(c) 3999
(d) None of these

Answer 1

Find the expression for ${{n}^{th}}$ of the expression in the L.H.S by finding the pattern of the terms present there. The pattern will be $\left( {{n}^{2}}-\left( n-1 \right) \right).\left( n-1 \right)!$. Write all the terms in this pattern and break the terms and write each term as ${{n}^{2}}.\left( n-1 \right)!-\left( n-1 \right).\left( n-1 \right)!$. Leave the term ${{n}^{2}}.\left( n-1 \right)!$ as it is and write $\left( n-1 \right).\left( n-1 \right)!={{\left( n-1 \right)}^{2}}.\left( n-2 \right)!$ by using the formula $p!=p.\left( p-1 \right)!$. Now, cancel the like terms and equate with the expression in the R.H.S to compare the value of n to get the answer.

Complete step by step answer:
Here we have been provided with the expression $1.\left( 0! \right)+3.\left( 1! \right)+7.\left( 2! \right)+13.\left( 3! \right)+21.\left( 4! \right)......$ up to n terms = $\left( 4000 \right)4000!$ and we are asked to find the value of n.
Now, let us consider the L.H.S, so we have,
$\Rightarrow L.H.S=1.\left( 0! \right)+3.\left( 1! \right)+7.\left( 2! \right)+13.\left( 3! \right)+21.\left( 4! \right)......$
We can write the above expression as:

& \Rightarrow L.H.S=\left( {{1}^{2}}-\left( 1-1 \right) \right).\left( 1-1 \right)!+\left( {{2}^{2}}-\left( 2-1 \right) \right).\left( 2-1 \right)!+\left( {{3}^{2}}-\left( 3-1 \right) \right).\left( 3-1 \right)! \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( {{4}^{2}}-\left( 4-1 \right) \right).\left( 4-1 \right)!+\left( {{5}^{2}}-\left( 5-1 \right) \right).\left( 5-1 \right)!...... \\\ \end{aligned}$$ On observing this pattern we can write the expression for the ${{n}^{th}}$ term, since we have started from $0!$ so the ${{n}^{th}}$ term will have $\left( n-1 \right)!$. Therefore we get the expression as: $$\begin{aligned} & \Rightarrow L.H.S=\left( {{1}^{2}}-\left( 1-1 \right) \right).\left( 1-1 \right)!+\left( {{2}^{2}}-\left( 2-1 \right) \right).\left( 2-1 \right)!+\left( {{3}^{2}}-\left( 3-1 \right) \right).\left( 3-1 \right)! \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( {{4}^{2}}-\left( 4-1 \right) \right).\left( 4-1 \right)!+\left( {{5}^{2}}-\left( 5-1 \right) \right).\left( 5-1 \right)!.+..... \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( {{n}^{2}}-\left( n-1 \right) \right).\left( n-1 \right)! \\\ \end{aligned}$$ Breaking the terms we get, $$\begin{aligned} & \Rightarrow L.H.S={{1}^{2}}.\left( 1-1 \right)!-\left( 1-1 \right).\left( 1-1 \right)!+{{2}^{2}}.\left( 2-1 \right)!-\left( 2-1 \right).\left( 2-1 \right)!+{{3}^{2}}.\left( 3-1 \right)!-\left( 3-1 \right).\left( 3-1 \right)! \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{{4}^{2}}.\left( 4-1 \right)!-\left( 4-1 \right).\left( 4-1 \right)!+{{5}^{2}}.\left( 5-1 \right)!-\left( 5-1 \right).\left( 5-1 \right)! \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+.....+{{n}^{2}}.\left( n-1 \right)!-\left( n-1 \right).\left( n-1 \right)! \\\ & \Rightarrow L.H.S={{1}^{2}}.0!-\left( 0 \right).0!+{{2}^{2}}.1!-\left( 1 \right).1!+{{3}^{2}}.2!-\left( 2 \right).2! \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{{4}^{2}}.3!-\left( 3 \right).3!+{{5}^{2}}.4!-\left( 4 \right).4! \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+.....+{{n}^{2}}.\left( n-1 \right)!-\left( n-1 \right).\left( n-1 \right)! \\\ \end{aligned}$$ Using the formula $p!=p.\left( p-1 \right)!$ we get, $$\begin{aligned} & \Rightarrow L.H.S={{1}^{2}}.0!-\left( 0 \right).0!+{{2}^{2}}.1!-{{1}^{2}}.0!+{{3}^{2}}.2!-{{2}^{2}}.1! \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{{4}^{2}}.3!-{{2}^{2}}.2!+{{5}^{2}}.4!-{{4}^{2}}.3! \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+.....+{{n}^{2}}.\left( n-1 \right)!-{{\left( n-1 \right)}^{2}}.\left( n-2 \right)! \\\ \end{aligned}$$ Cancelling the like terms we will be left with the expression: $$\begin{aligned} & \Rightarrow L.H.S=-0.0!+{{n}^{2}}.\left( n-1 \right)! \\\ & \Rightarrow L.H.S={{n}^{2}}.\left( n-1 \right)! \\\ \end{aligned}$$ Equating the above expression with the given R.H.S we get, $$\begin{aligned} & \Rightarrow {{n}^{2}}.\left( n-1 \right)!=\left( 4000 \right)4000! \\\ & \Rightarrow n.n.\left( n-1 \right)!=\left( 4000 \right)4000! \\\ \end{aligned}$$ We know that $n.\left( n-1 \right)!=n!$, so we get, $$\therefore n.n!=\left( 4000 \right)4000!$$ On comparing both the sides we can clearly see that $n=4000$. **So, the correct answer is “Option a”.** **Note:** Note that we cannot add the terms by simplifying them individually as there are n terms and we don’t know the value of n. Even if we know the value of n = 4000 then also we cannot find the factorial of such a large number and that is why we need to form a series for the simplification. Similar types of series also appear in chapters like sequence and series, inverse trigonometry etc. So it is necessary to obtain the required pattern to solve these questions.