If ∑k=1m(k2+1)k!=1999(2000!) then m = _____________

∑k=1m(k2+1)k!=∑k=1m[(k+1)2−2k]k! =∑k=1m[(k+1)2k!−2k.k!]=∑k=1m(k+1)(k+1)k!−2∑k=1mk k! =∑k=1m(k+1)(k+1)!−2∑k=1m{(k+1)!−k!}=∑k=1m{(k+2)−1}(k+1)!−2∑k=1m{(k+1)−1}k! =∑k=1m[(k+2)!(k+1)!]−2∑k=1m[(k+1)!−k!] =[(3i−2!)+(4!−3!)+(5!−4!)+..\+{(m+2)!−(m+1)}] −2[(2!−1)+(3!−2!)+(3!−3!)+\..\+{(m+1)!−m!}] =[(m+2)!−2!]−2[(m+1)!−1!] =(m+2)!−2−2(m+1)1+2 =(m+2)(m+1)!−2(m+1)!=(m+2−2)(m+1)!=m(m+1)! ∴ m(m+1)!=1999 (2000!) ⇒ m=1999

If ∑k=1m(k2+1)k!=1999(2000!) then m = _____________

∑k=1m(k2+1)k!=∑k=1m[(k+1)2−2k]k! =∑k=1m[(k+1)2k!−2k.k!]=∑k=1m(k+1)(k+1)k!−2∑k=1mk k! =∑k=1m(k+1)(k+1)!−2∑k=1m{(k+1)!−k!}=∑k=1m{(k+2)−1}(k+1)!−2∑k=1m{(k+1)−1}k! =∑k=1m[(k+2)!(k+1)!]−2∑k=1m[(k+1)!−k!] =[(3i−2!)+(4!−3!)+(5!−4!)+..\+{(m+2)!−(m+1)}] −2[(2!−1)+(3!−2!)+(3!−3!)+\..\+{(m+1)!−m!}] =[(m+2)!−2!]−2[(m+1)!−1!] =(m+2)!−2−2(m+1)1+2 =(m+2)(m+1)!−2(m+1)!=(m+2−2)(m+1)!=m(m+1)! ∴ m(m+1)!=1999 (2000!) ⇒ m=1999

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If $\sum_{k = 1}^{m} (k^{2} + 1) k! = 1999 (2000!)$ then m = _____________

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2000

1999

2001

2002

answer is B.

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Detailed Solution

$\sum_{k = 1}^{m} (k^{2} + 1) k! = \sum_{k = 1}^{m} [{(k + 1)}^{2} - 2 k] k! = \sum_{k = 1}^{m} [{(k + 1)}^{2} k! - 2 k . k!] = \sum_{k = 1}^{m} (k + 1) (k + 1) k! - 2 \sum_{k = 1}^{m} k k! = \sum_{k = 1}^{m} (k + 1) (k + 1)! - 2 \sum_{k = 1}^{m} {(k + 1)! - k!} = \sum_{k = 1}^{m} {(k + 2) - 1} (k + 1)! - 2 \sum_{k = 1}^{m} {(k + 1) - 1} k! = \sum_{k = 1}^{m} [(k + 2)! (k + 1)!] - 2 \sum_{k = 1}^{m} [(k + 1)! - k!] = [(3 i - 2!) + (4! - 3!) + (5! - 4!) + .. \ + {(m + 2)! - (m + 1)}] - 2 [(2! - 1) + (3! - 2!) + (3! - 3!) + \ .. \ + {(m + 1)! - m!}] = [(m + 2)! - 2!] - 2 [(m + 1)! - 1!] = (m + 2)! - 2 - 2 (m + 1) 1 + 2 = (m + 2) (m + 1)! - 2 (m + 1)! = (m + 2 - 2) (m + 1)! = m (m + 1)! ∴ m (m + 1)! = 1999 (2000!) \Rightarrow m = 1999$