The given sum $1 \times 1! + 2 \times 2! + ............. + 50 \times 50!$ is equal to
(a) $51!$
(b) $51! - 1$
(c) $51! + 1$
(d) $2 \times 51!$

Hint: In this problem use some basic properties of factorials and rearrange the terms to get a desired answer.

We have to find the sum of $1 \times 1! + 2 \times 2! + ............. + 50 \times 50!$
This can be rewritten as
\[\left( {2 - 1} \right)1! + \left( {3 - 1} \right)2! + \left( {4 - 1} \right)3! + ...............................\left( {50 - 1} \right)49! + \left( {51 - 1} \right)50!\]

Separating the positive terms and negative terms, we get
\[\left( {2 \times 1! + 3 \times 2! + 4 \times 3! + ..............50 \times 49! + 51 \times 50!} \right) - \left({1! + 2! + 3! + ..............49! + 50!} \right)\]

which can be written as
\[\left( {2! + 3! + 4! + ..............50! + 51!} \right) - \left( {1! + 2! + 3! + ..............49! + 50!} \right)\]

Adding and subtracting \[1\] we get
\[\left[ {\left( {1! + 2! + 3! + ..............49! + 50! + 51!} \right) - \left( {1! + 2! + 3! + ..............49! + 50!}\right)} \right] - 1\]
Cancelling the common terms, we will get
\[51! - 1\]
Thus the answer is option (b) $51! - 1$

Note: In this type of problems we can also solve by the summation method by rewriting the equation and using the formula $\sum\limits_{n = 1}^n {\left( {n + 1} \right)! - n! = \left( {n + 1} \right)! - 1}$ directly.