For all $\mathrm{n}\in \mathrm{N},1\times 1!+2\times 2!+3\times 3!+\dots +\mathrm{n}\times \mathrm{n}!$ is equal to

detailed solution

Correct option is C

Let the statement P(n) be defined asP(n):1×1!+2×2!+3×3!…+n×n!=(n+1)!−1 for all natural numbers n. for all natural numbers n. P(1):1×1!=1=2−1=2!−1Assume that P(n) is true for some natural number k, i.eP(k):1×1!+2×2!+3×3!+….+k×k!=(k+1)!−1---------(i)To prove P(k + 1) is true, we haveP(k+1):1×1!+2×2!+3×3!+…+k×k!+(k+1)×(k+1)!=(k+1)!−1+(k+1)!×(k+1)=(k+1+1)(k+1)!−1=(k+2)(k+1)!−1=(k+2)!−1Thus, P(k + 1) is true, whenever P(k) is true. Therefore, by the principle of mathematical induction,P(n) is true for all natural numbers n.