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Question: The given sum \(1 \times 1! + 2 \times 2! + ............. + 50 \times 50!\) is equal to (a) \(51!\...

The given sum 1×1!+2×2!+.............+50×50!1 \times 1! + 2 \times 2! + ............. + 50 \times 50! is equal to
(a) 51!51!
(b) 51!−151! - 1
(c) 51!+151! + 1
(d) 2×51!2 \times 51!

Explanation

Solution

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 1×1!+2×2!+.............+50×50!1 \times 1! + 2 \times 2! + ............. + 50 \times 50!
This can be rewritten as
(2−1)1!+(3−1)2!+(4−1)3!+...............................(50−1)49!+(51−1)50!\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
(2×1!+3×2!+4×3!+..............50×49!+51×50!)−(1!+2!+3!+..............49!+50!)\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
(2!+3!+4!+..............50!+51!)−(1!+2!+3!+..............49!+50!)\left( {2! + 3! + 4! + ..............50! + 51!} \right) - \left( {1! + 2! + 3! + ..............49! + 50!} \right)

Adding and subtracting 11 we get
[(1!+2!+3!+..............49!+50!+51!)−(1!+2!+3!+..............49!+50!)]−1\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!−151! - 1
Thus the answer is option (b) 51!−151! - 1

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