Simple Series

Calculus Level 3

For sequence { a n } \left\{ { a }_{ n } \right\} , the following holds: a 1 = 5 { a }_{ 1 }=5 and a 100 = 200 { a }_{ 100 }=200 . Evaluate ∑ k = 1 99 a k − ∑ k = 1 50 a 2 k − ∑ k = 1 49 a 2 k + 1 \sum _{ k=1 }^{ 99 }{ { a }_{ k } } -\sum _{ k=1 }^{ 50 }{ { a }_{ 2k } } -\sum _{ k=1 }^{ 49 }{ { a }_{ 2k+1 } }


The answer is -195.

Min-woo Lee by Min-woo Lee , 23, South Korea

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1 solution

Min-woo Lee
Jul 31, 2014

∑ k = 1 99 a k \sum _{ k=1 }^{ 99 }{ { a }_{ k } } is sum of sequence { a n } \left\{ { a }_{ n } \right\} from one to 99. Since subtracting ∑ k = 1 50 a 2 k \sum _{ k=1 }^{ 50 }{ { a }_{ 2k } } will take away even numbered terms and a 100 { a }_{ 100 } , it gives you [sum of all odd numbered terms - a_100]. Subtracting ∑ k = 1 49 a 2 k + 1 \sum _{ k=1 }^{ 49 }{ { a }_{ 2k+1 } } will take away all odd numbered terms from 3 to 99, ergo we are left with a 1 − a 100 { a }_{ 1 } - { a }_{ 100 } . 5 − 200 = − 195 \boxed{5 - 200 = -195}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Level 3 is not deserving!!! Sorry!!

Kartik Sharma - 6 years, 10 months ago

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