1. Tardigrade
  2. Question
  3. Mathematics
  4. The statement P(r)=∑r=1n r(r !)=(n+1) !-1 is true for

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

Principle of Mathematical Induction

Solution:

Let $P(k)$ is true. $ P(k)=\sum_{k=1}^{n} k(k !) $ (Note $: k=1, \ldots r$ or, $k=1 \ldots n$ 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, \ldots 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 $\forall n \in N$