Is this a Progression

Algebra Level 3

Let a 1 , a 2 , a 3 , . . . a n { a }_{ 1 },{ a }_{ 2 }, a_3, ... { a }_{ n } be real numbers such that

a 1 + a 2 − 1 + a 3 − 2 + . . . + a n − ( n − 1 ) = 1 2 ( a 1 + a 2 + . . . + a n ) − n ( n − 3 ) 4 \sqrt { { a }_{ 1 } } +\sqrt { { a }_{ 2 }-{ 1 } } +\sqrt { { a }_{ 3 }-{ 2 } } +...+\sqrt { { a }_{ n }-{ \left( n-1 \right) } } = \frac { 1 }{ 2 } \left( { a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ n } \right) -\frac { { n }{ \left( n-3 \right) } }{ 4 }

Then compute the value of ∑ i = 1 100 a i \displaystyle \sum _{ i=1 }^{ 100 }{ { a }_{ i } } .


The answer is 5050.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Chew-Seong Cheong
Mar 15, 2018

The first few i ∈ N i \in \mathbb N appears that a i = i a_i = i . Assume that it is true then we have:

{ L H S = ∑ i = 1 n a i − ( i − 1 ) = ∑ i = 1 n i − ( i − 1 ) = ∑ i = 1 n 1 = n R H S = 1 2 ∑ i = 1 n a i − n ( n − 3 ) 4 = 1 2 ∑ i = 1 n i − n ( n − 3 ) 4 = 1 2 ( n ( n + 1 ) 2 ) − n ( n − 3 ) 4 = n \begin{cases} \displaystyle LHS = \sum_{i=1}^n \sqrt{a_i-(i-1)} = \sum_{i=1}^n \sqrt{i-(i-1)} = \sum_{i=1}^n 1 = n \\ \displaystyle RHS = \frac 12 \sum_{i=1}^n a_i - \frac {n(n-3)}4 = \frac 12 \sum_{i=1}^n i - \frac {n(n-3)}4 = \frac 12\left(\frac {n(n+1)}2\right) - \frac {n(n-3)}4 = n \end{cases}

Therefore L H S = R H S LHS = RHS , implying that the claim a i = i a_i = i is true. And ∑ i = 1 100 a i = ∑ i = 1 100 i = 100 ( 100 + 1 ) 2 = 5050 \displaystyle \sum_{i=1}^{100} a_i = \sum_{i=1}^{100} i = \frac {100(100+1)}2 = \boxed{5050} .

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