Q.

For all n∈N,1×1!+2×2!+3×3!+…+n×n! is equal to

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a

(n+1)!−2

b

(n+1)!

c

(n+1)!-1

d

(n+1)!-3

answer is C.

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Detailed Solution

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.
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