Find The Sum

Algebra Level 4

If 1 3 − 2 2 1 ! + 2 3 − 3 2 2 ! + 3 3 − 4 2 3 ! + ⋯ + 10 0 3 − 10 1 2 100 ! = 1 − a b \dfrac{1^3 - 2^2}{1!} + \dfrac{2^3 - 3^2}{2!} + \dfrac{3^3 - 4^2}{3!} + \cdots + \dfrac{100^3 - 101^2}{100!} = 1 - \dfrac{a}{b} for some coprime positive integers a , b , a, b, find the last three digits of a . a.


The answer is 201.

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Sreejato Bhattacharya by Sreejato Bhattacharya , 22, India

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1 solution

n 3 − ( n + 1 ) 2 n ! = n 3 n ! − ( n + 1 ) 2 n ! = n 2 ( n − 1 ) ! − ( n + 1 ) 2 n ! \frac{n^3-(n+1)^2}{n!} = \frac{n^3}{n!} - \frac{(n+1)^2}{n!} = \frac{n^2}{(n-1)!} -\frac{(n+1)^2}{n!}

1 3 − 2 2 1 ! + 2 3 − 3 2 2 ! + 3 3 − 4 2 3 ! + . . . . . . . . + 10 0 3 − 10 1 2 100 ! \frac{1^3-2^2}{1!} + \frac{2^3-3^2}{2!} + \frac{3^3-4^2}{3!} + ........+\frac{100^3-101^2}{100!}

= ( 1 3 1 ! − 2 2 1 ! ) + ( 2 2 1 ! − 3 2 2 ! ) + ( 3 2 2 ! − 4 2 3 ! ) + . . . . . . . . . . + ( 10 0 2 99 ! − 10 1 2 100 ! ) = (\frac{1^3}{1!} -\frac{2^2}{1!}) + (\frac{2^2}{1!} -\frac{3^2}{2!}) + (\frac{3^2}{2!} -\frac{4^2}{3!}) + .......... + (\frac{100^2}{99!} -\frac{101^2}{100!})

= 1 − 10 1 2 100 ! = 1-\frac{101^2}{100!}

⟶ a b = 10 1 2 100 ! \longrightarrow \frac{a}{b} = \frac{101^2}{100!}

a = 10 1 2 = 10201 a = 101^2 = 10201 . Last three digits 201 \boxed{201}

17 Helpful 0 Interesting 0 Brilliant 0 Confused

One may transform n 3 − ( n + 1 ) 2 n ! \dfrac{n^3 - (n+1)^2}{n!} to n 3 n ! − ( n + 1 ) 2 n ! = n 3 n ! − ( n + 1 ) 3 ( n + 1 ) ! = f ( n ) − f ( n + 1 ) \dfrac{n^3}{n!} - \dfrac{(n+1)^2}{n!} = \dfrac{n^3}{n!} - \dfrac{(n+1)^3}{(n+1)!} = f(n) - f(n+1) where f ( n ) = n 3 n ! f(n) = \dfrac{n^3}{n!} . This turns a telescopic which ultimately gives f ( 1 ) − f ( 101 ) f(1) - f(101) . Though the above solution is quite similar as this.

Sagnik Saha - 7 years, 1 month ago

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I did it your way.

Anish Puthuraya - 7 years, 1 month ago

same here

Anirudha Nayak - 7 years, 1 month ago

oh ..it was so easy....

Max B - 7 years, 1 month ago

Precisely my intended solution. Good job!

Sreejato Bhattacharya - 7 years, 1 month ago

wow i also followed the same method bu kept in the x form

Anish Kelkar - 7 years ago

Done the same way !

Dinesh Chavan - 7 years, 1 month ago

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Here's a good challenge. Create a telescoping series where the answer is NOT the first term minus the last (assuming all terms are in order)

Trevor Arashiro - 6 years, 8 months ago

done d same way!

Pradeep Ch - 7 years, 1 month ago

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