Question

What is the sum of the first \(n\)-terms of the series, whose \(k\)-term is \(k!\times k\)?

• A. \((n+1)!^{\,n}-1\)
• B. \((n+1)^n-1\)
• C. \((n+1)!-1\)
• D. \(3n-2\)

Hint:

For factorial series, try to write the term as a difference of two consecutive factorials. This often creates a telescoping series.

Answer 1

The Correct Option is C

Step 1: Write the \(k\)-th term.
Given \(k\)-th term is \[ k!\times k \] That is, \[ T_k=k\cdot k! \]

Step 2: Express \(k\cdot k!\) in telescopic form.
We know that \[ (k+1)!=(k+1)k! \] So, \[ (k+1)!-k!=(k+1)k!-k! \] \[ =kk! \] Therefore, \[ k\cdot k!=(k+1)!-k! \]

Step 3: Sum the first \(n\) terms.
\[ S_n=\sum_{k=1}^{n} k\cdot k! \] \[ =\sum_{k=1}^{n}\left((k+1)!-k!\right) \] \[ =(2!-1!)+(3!-2!)+(4!-3!)+\cdots+((n+1)!-n!) \]

Step 4: Cancel the common terms.
Most terms cancel out, so \[ S_n=(n+1)!-1! \] Since \[ 1!=1, \] we get \[ S_n=(n+1)!-1. \]

Step 5: Final conclusion.
Therefore, the required sum is \[ \boxed{(n+1)!-1} \]