Easy one

Calculus Level 4

10 0 2 100 ! + ∑ k = 1 100 ( k 2 − 3 k + 1 ) 1 ( k − 1 ) ! = ? \displaystyle \dfrac{100^{2}}{100!} + \sum_{k=1}^{100} ( k^{2} - 3k + 1)\dfrac{1}{(k - 1)!} = \ ?


The answer is 0.

U Z by U Z , 24, Guyana

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1 solution

Chew-Seong Cheong
Dec 22, 2014

∑ k = 1 100 k 2 − 3 k + 1 ( k − 1 ) ! = ∑ k = 1 100 ( k − 1 ) 2 − k ( k − 1 ) ! = ∑ k = 1 100 ( k − 1 ( k − 2 ) ! − k ( k − 1 ) ! ) \displaystyle \sum _{k=1} ^{100} {\dfrac {k^2-3k+1} {(k-1)!}} = \sum _{k=1} ^{100} {\dfrac {(k-1)^2-k} {(k-1)!}} = \sum _{k=1} ^{100} {\left( \dfrac {k-1} {(k-2)! } - \dfrac {k}{(k-1)!} \right)}

= 1 2 − 3 ˙ 1 + 1 0 ! + ∑ k = 2 100 ( k − 1 ( k − 2 ) ! − k ( k − 1 ) ! ) \displaystyle = \dfrac {1^2-3\dot{}1 +1}{0!} + \sum _{k=2} ^{100} {\left( \dfrac {k-1} {(k-2)! } - \dfrac {k}{(k-1)!} \right)}

= − 1 0 ! + 1 0 ! − 2 1 ! + 2 1 ! − 3 2 ! + 3 2 ! − 4 3 ! + . . . + 99 98 ! − 100 99 ! = − 100 99 ! = - \dfrac {1}{0!} + \dfrac {1}{0!} - \dfrac {2}{1!} + \dfrac {2}{1!} - \dfrac {3}{2!} + \dfrac {3}{2!} - \dfrac {4}{3!} + ... + \dfrac {99}{98!} - \dfrac {100}{99!} = - \dfrac {100}{99!}

Therefore,

10 0 2 100 ! + ∑ k = 1 100 k 2 − 3 k + 1 ( k − 1 ) ! = 10 0 2 100 ! − 100 99 ! = 100 99 ! − 100 99 ! = 0 \displaystyle \dfrac {100^2}{100!} + \sum _{k=1} ^{100} {\dfrac {k^2-3k+1} {(k-1)!}} = \dfrac {100^2}{100!} - \dfrac {100}{99!} = \dfrac {100}{99!} - \dfrac {100}{99!} = \boxed{0}

15 Helpful 1 Interesting 1 Brilliant 0 Confused

Nice solution Sir

∑ k = 1 100 k 2 − 3 k + 1 ( k − 1 ) ! = ∑ k = 1 100 ( k − 1 ) 2 − k ( k − 1 ) ! = ∑ k = 1 100 ( k − 1 ) 2 ( k − 1 ) ! − k 2 ( k ) ! \displaystyle \sum _{k=1} ^{100} {\dfrac {k^2-3k+1} {(k-1)!}} = \sum _{k=1} ^{100} {\dfrac {(k-1)^2-k} {(k-1)!}} = \sum _{k=1} ^{100} \dfrac{(k - 1)^{2}}{(k - 1)!} - \dfrac{k^{2}}{(k)!}

U Z - 6 years, 5 months ago

You can also factorise the quadratic like (k-1)(k-2)-1 this will reduces the whole thing by forming a telescopic series :P Enjoy

akash omble - 6 years, 5 months ago

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