n = 2 ∑ 2 0 1 7 k = 2 ∏ n ( 1 + k 1 ) n 1
The above expression can be expressed in the form b a , where a and b are coprime positive integers. Find a + b .
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Every Problem I solve and when I come to the solutions I see only 2 words everytime on the top: Rishabh Cool.
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lol,very true @Rishabh Cool does write beautiful solutions,keep up the good work! :)
Exactly. The guys got something I have to say
Do u use latex editor(which one?)?
n = 2 ∑ 2 0 1 7 k = 2 ∏ n ( 1 + k 1 ) n 1 = n = 2 ∑ 2 0 1 7 k = 2 ∏ n ( k k + 1 ) n 1 = n = 2 ∑ 2 0 1 7 2 n ! ( n + 1 ) ! n 1 = n = 2 ∑ 2 0 1 7 n ( n + 1 ) ! 2 n ! = 2 n = 2 ∑ 2 0 1 7 ( n + 1 ) ! ( n − 1 ) ! = 2 n = 2 ∑ 2 0 1 7 ( n − 1 ) ! ( n − 1 ) ! ( n ( n + 1 ) 1 ) = 2 n = 2 ∑ 2 0 1 7 ( n 1 − n + 1 1 ) = 2 ( n = 2 ∑ 2 0 1 7 n 1 − n = 3 ∑ 2 0 1 8 n 1 ) = 2 ( 2 1 − 2 0 1 8 1 ) = 1 − 1 0 0 9 1 = 1 0 0 9 1 0 0 8
⇒ a + b = 1 0 0 8 + 1 0 0 9 = 2 0 1 7
k = 2 ∏ n 1 + k 1 = k = 2 ∏ n k k + 1 = 2 3 × 3 4 × 4 5 × ⋯ × n − 1 n × n n + 1 = 2 n + 1
Substituting this in given summation ,
n = 2 ∑ 2 0 1 7 2 n + 1 n 1 = n = 2 ∑ 2 0 1 7 n ( n + 1 ) 2 = 2 n = 2 ∑ 2 0 1 7 n ( n + 1 ) ( n + 1 ) − n = 2 n = 2 ∑ 2 0 1 7 n 1 − n + 1 1 = 2 [ 2 1 − 3 1 + 3 1 − 4 1 + 4 1 − 5 1 ⋯ − 2 0 1 7 1 + 2 0 1 7 1 − 2 0 1 8 1 ] = 2 [ 2 1 − 2 0 1 8 1 ] = 2 × 2 1 [ 1 − 1 0 0 9 1 ] = 1 0 0 9 1 0 0 9 − 1 = 1 0 0 9 1 0 0 8
Hence a = 1 0 0 8 , b = 1 0 0 9 ⇒ a + b = 1 0 0 8 + 1 0 0 9 = 2 0 1 7 .
Great explanation of the telescoping product and sum.
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Let Q denote the given expression. k = 2 ∏ n ( 1 + k 1 ) = k = 2 ∏ n ( k k + 1 ) = 2 3 × 3 4 × 4 5 × ⋯ × n n + 1 = 2 n + 1 Hence , Q = n = 2 ∑ 2 0 1 7 k = 2 ∏ n ( 1 + k 1 ) n 1 = n = 2 ∑ 2 0 1 7 n ( n + 1 ) 2 = 2 n = 2 ∑ 2 0 1 7 ( n 1 − n + 1 1 ) ( T e l e s c o p i c S e r i e s ) = 2 ( 2 1 − 2 0 1 8 1 ) = 1 0 0 9 1 0 0 8 ∴ 1 0 0 8 + 1 0 0 9 = 2017