Permutation-Combination
Aptitude

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Q.

What is the value of $1×1! + 2×2! + 3×3! + ............ n×n!$;

where $n!$ means $n$ factorial or $n(n-1)(n-2)...1$

 A.

$n(n-1)(n-1)!$

 B.

${(n+1)!}/{n(n-1)}$

 C.

$(n+1)! - n!$

 D.

$(n + 1)! - 1!$

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Solution:
Option(D) is correct

\begin{align*}
1×1!\\ 
&= (2 -1)×1!\\
&= 2×1! - 1×1! \\
&= 2! - 1!\\
2×2!\\ 
&= (3 - 1)×2!\\ 
&= 3×2! - 2!\\
& = 3! - 2!\\
3×3!\\ 
&= (4 - 1)×3! \\
&= 4×3! - 3! \\
&= 4! - 3!\\
&\cdots\\
&\cdots\\
&\cdots\\
n×n! \\
&= (n+1 - 1)×n! \\
&= (n+1)(n!) - n! \\
&= (n+1)! - n!
\end{align*}

Summing up all these terms, we get $(n+1)! - 1!$

Edit: Thank you david for pointing out the error. Corrected the question.


(5) Comment(s)


Saim
 ()

add and subtract 1 to each term. you get the answer within 30 secs :)



Sai
 ()

simply verify options



David
 ()

There is a mistake in the question...

it should be $3 \times 3!$ instead of $3! \times 3!$


Deepak
 ()

Thank you for letting me know the error. Corrected


PRATYUSH ANAND
 ()

Really Good+ Question Smile