The sum
2 1 + 1 2 1 + 3 2 + 2 3 1 + 4 3 + 3 4 1 + ⋯ + 1 0 0 9 9 + 9 9 1 0 0 1
can be expressed as b a , where a and b are coprime positive integers.
What is the value of a + b ?
This problem is adapted from a past year KMC Contest.
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super
Only 4th step is wrong
Good One. By the way, did you use latex?
what do you call the sign that looks like letter "E"?
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Sigma? Σ <- This one?
Its used to signify summation.
http://en.wikipedia.org/wiki/%E2%88%91
By rationalizing the denominator of each term we can get the series equal to 1- 1/sqrt(2)+ 1/sqrt(2)- 1/sqrt(3)+ 1/sqrt(3)- 1/sqrt(4) +1/sqrt(4)+...-1/sqrt(99)+1/sqrt(99)-1/sqrt(100) = 1-1/10= 9/10 and so a+b= 9+10=19!
This one's easier
{1/(2root1+1root2)}+{1/(3root2+2root3)}+{1/(4root3+3root4)}+------------+{1/(100root99+99root100)}=[1/root2root1(root2+root1)]+[1/root3root2(root3+root2)]+[1/root4root3(rot4+root3)]+-----------+[1/root100root99(root100+root99)]
after rationalizing we get
common terms cancels out and we left with
1-(1/root2)+(1/root2)-(1/root3)+(1/root3)-(1/root4)+(1/root4)-------------------(1/root99)+(1/root99)-(1/root100)=(1-1/1root100)=1-(1/10)=9/10=a/b.
so, a=9, b=10, a+b=9+10=19. Is the required answer.
Nice solution, but maybe you should use Latex next time.
rationalize each term and split them as single term we get sqrt1-1/sqrt2+1/sqrt2-1/sqrt3+...........+1/sqrt99-1/sqrt100 which is equal to 1-1/10=9/10. a+b=9+10=19
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2 1 + 1 2 1 + 3 2 + 2 3 1 + … + 3 2 + 2 3 1
= n = 1 ∑ 9 9 n n + 1 + ( n + 1 ) n 1
= n = 1 ∑ 9 9 ( n ) ( n + 1 ) ( n + n + 1 ) 1
= n = 1 ∑ 9 9 ( n ) ( n + 1 ) ( n + n + 1 ) 1 × n + 1 − n n + 1 − n
= n = 1 ∑ 9 9 ( n ) ( n + 1 ) ( ( n + 1 ) 2 − ( n ) 2 ) n + 1 − n
= n = 1 ∑ 9 9 ( n ) ( n + 1 ) n + 1 − n
= n = 1 ∑ 9 9 n 1 − n + 1 1
= ( 1 1 − 2 1 ) + ( 2 1 − 3 1 ) + … + ( 9 9 1 − 1 0 0 1 )
= 1 1 − 1 0 0 1
= 1 − 1 0 1
= 1 0 9
Therefore, a = 9 , b = 1 0 , and a + b = 1 9