Question

The value of $(3\cdot 2\cdot {}^1{P}_0-4\cdot 3\cdot {}^2{P}_1+5\cdot 4\cdot {}^3{P}_2-\cdots \text{upto}101\text{terms})$ $+(2!-3!+4!-\cdots \text{upto}101\text{terms})$ is equal to

• A. $2!-102!$
• B. $2!-103!$
• C. $2!+103!$
• D. $3!+103!$

Answer 1

The correct option is C $2!+103!$

Let
$S=(3\cdot 2\cdot {}^1{P}_0-4\cdot 3\cdot {}^2{P}_1+5\cdot 4\cdot {}^3{P}_2-\cdots \text{upto}101\text{terms})+(2!-3!+4!-\cdots \text{upto}101\text{terms})$
$\Rightarrow S=(3\cdot 2\cdot {}^1{P}_0-4\cdot 3\cdot {}^2{P}_1+5\cdot 4\cdot {}^3{P}_2-\cdots +103\cdot 102\cdot {}^{101}{P}_{100})+(2!-3!+4!-\cdots +102!)$

Let
$S_1=(3\cdot 2\cdot {}^1{P}_0-4\cdot 3\cdot {}^2{P}_1+5\cdot 4\cdot {}^3{P}_2-\cdots \text{upto}101\text{terms})$
$\Rightarrow S_1=3\cdot 2\cdot \frac{1!}{1!}-4\cdot 3\cdot \frac{2!}{1!}+5\cdot 4\cdot \frac{3!}{1!}-\cdots +103\cdot 102\cdot \frac{101!}{1!}$
$\Rightarrow S_1=3\cdot 2\cdot 1!-4\cdot 3\cdot 2!+5\cdot 4\cdot 3!-\cdots +103\cdot 102\cdot 101!$
$\Rightarrow S_1=3!-4!+5!-\cdots +103!$

Putting the value of $S_1$ in $S$
$\therefore S=(3!-4!+5!-\cdots +103!)+(2!-3!+4!-5!+\cdots +102!)$
$\Rightarrow S=2!+103!$