Question

The statement $P(r)={\sum }_{r=1}^{n}r(r!)=(n+1)!-1$ is true for

• A. $r>2$
• B. $\forall r\in N$
• C. for no value of $r$
• D. $r>1$

Answer 1

Answer: (B)

Let $P(k)$ is true. $P(k)={\sum }_{k=1}^{n}k(k!)$ (Note $:k=1,\dots r$ or, $k=1\dots n$ have same meaning as we have consider $r,n\in N$ ) $\begin{array}{l}={\sum }_{k=1}^n(k+1-1)(k!)\\ ={\sum }_{k=1}^n(k+1)!-{\sum }_{k=1}^n(k!)\text{ for }k=n,n-1,\dots 1\\ =((n+1)!-n!+n!-(n-1)!+(n-1)!-(n-2)!+3!-2!+2!-1!)\\ \therefore P(k)=(n+1)!=1!\end{array}$ Short Cut Method: It is enough if we proceed as follows. when $n=1$, L.H. $S.=1\cdot 1!=2!-1!$ when $n=2$, L. H.S. $=1\cdot 1!+2\cdot 2!=3!-1$ when $n=3$, L. H.S. $=1\cdot 1!+2\cdot 2!+3\cdot 3!=4!-1$ for $n=n$, L. H.S. $=(n+1)!-1$ Hence, $P(n)$ is true $\forall n\in N$