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.