Question

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

Answer 1

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 $\sum _{n=1}^n\left(n+1\right)!-n!=\left(n+1\right)!-1$ directly.