Question

The value of $^2{P_1}{ + ^3}{P_1} + ........{ + ^n}{P_1}$ is equal to
A. $\dfrac{{{n^2} - n + 2}}{2}$
B. $\dfrac{{{n^2} + n + 2}}{2}$
C. $\dfrac{{{n^2} + n - 1}}{2}$
D. $\dfrac{{{n^2} - n - 1}}{2}$
E. $\dfrac{{{n^2} + n - 2}}{2}$

Answer 1

Hint : Add and subtract $1$ in the value given. Use the basic formulas of Permutation that is $^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$. Division type of permutation like $^7{P_5} = \dfrac{{7!}}{{\left( {7 - 5} \right)!}} = \dfrac{{7!}}{{2!}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2!}}{{2!}} = 7 \times 6 \times 5 \times 4 \times 3 = 2520$
Just add the values, add them and get the results. The value of $^n{P_1}$ always gives $n$.Similarly for the other values too. Use of sum of series of natural numbers.

Complete step-by-step answer :
We are given with $^2{P_1}{ + ^3}{P_1} + ........{ + ^n}{P_1}$
From the permutation formula that is $^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$;
Now, Let’s see for first term $^2{P_1}$, put the values in the above formula we get:

$$^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}} \\\ ^2{P_1} = \dfrac{{2!}}{{\left( {2 - 1} \right)!}} = \dfrac{{2!}}{{1!}} = \dfrac{{2 \times 1}}{1} = 2 \\\ \;$$

Similarly check for $^3{P_1}$, and we get:

$$^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}} \\\ ^3{P_1} = \dfrac{{3!}}{{\left( {3 - 1} \right)!}} = \dfrac{{3!}}{{2!}} = \dfrac{{3 \times 2 \times 1}}{{2 \times 1}} = 3 \\\ \;$$

And, also for others we will get the same. So overall
$^n{P_1} = n$.
Now, the question becomes:

$$^2{P_1}{ + ^3}{P_1} + ........{ + ^n}{P_1} \\\ = 2 + 3 + 4 + ....... + n \;$$

Adding $1$ in the above value, but if we are adding one new value then we have to subtract that term to get the original term.
Now,

$$2 + 3 + 4 + ....... + n \\\ = 1 + 2 + 3 + 4 + ....... + n - 1 \;$$

Separate the values into two parts that is:
$(1 + 2 + 3 + 4 + ....... + n) - 1$ ……………(i)
From the Series we know that the sum of series of natural numbers up to $n$terms is:
$(1 + 2 + 3 + 4 + ....... + n) = \dfrac{{n(n + 1)}}{2}$
Put this value in (i) we get:

$$(1 + 2 + 3 + 4 + ....... + n) - 1 \\\ = \dfrac{{n(n + 1)}}{2} - 1 \;$$

Take the common LCM and solve further:
On further solving we get:

$$\dfrac{{n(n + 1)}}{2} - 1 \\\ = \dfrac{{{n^2} + n}}{2} - 1 \\\ = \dfrac{{{n^2} + n - 2}}{2} \;$$

Hence, the value of $^2{P_1}{ + ^3}{P_1} + ........{ + ^n}{P_1}$ is equal to $\dfrac{{{n^2} + n - 2}}{2}$.
Therefore, the correct option is Option E i.e$\dfrac{{{n^2} + n - 2}}{2}$.
So, the correct answer is “OPTION E”.

Note : Remember the formula of basic permutation and methods to solve it step by step.
I.Always solve step by step, don’t do the solving part at once, otherwise it would give a chance of error.
II.Always remember the formula of sum of series like sum of natural numbers, sum of squares of natural numbers, etc.