Series JEE 3

Algebra Level 4

∑ r = 1 50 [ 1 49 + r − 1 2 r ( 2 r − 1 ) ] = ? \large \displaystyle \sum_{r=1}^{50} \left [ \frac{1}{49+r}-\frac{1}{2r(2r-1)} \right]= \, ?


The answer is 0.01.

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Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Ronak Agarwal
Dec 31, 2015

Divide the sum into two parts :

S = ∑ r = 50 99 1 r − ∑ r = 1 50 1 2 r ( 2 r − 1 ) = T − U \displaystyle S=\sum _{ r=50 }^{ 99 }{ \frac { 1 }{ r } } -\sum _{ r=1 }^{ 50 }{ \frac { 1 }{ 2r(2r-1) } } =T-U

We concentrate on U U :

U = ∑ r = 1 50 1 2 r − 1 − 1 2 r = ∑ r = 1 50 1 2 r − 1 − ∑ r = 1 50 1 2 r \displaystyle U = \sum _{ r=1 }^{ 50 }{ \frac { 1 }{ 2r-1 } -\frac { 1 }{ 2r } } =\sum _{ r=1 }^{ 50 }{ \frac { 1 }{ 2r-1 } } -\sum _{ r=1 }^{ 50 }{ \frac { 1 }{ 2r } }

U = ∑ r = 1 50 1 2 r − 1 − ∑ r = 1 49 1 2 r − 1 100 \displaystyle U = \sum _{ r=1 }^{ 50 }{ \frac { 1 }{ 2r-1 } } -\sum _{ r=1 }^{ 49 }{ \frac { 1 }{ 2r } } -\frac { 1 }{ 100 }

⇒ U = ∑ r = 1 50 1 2 r − 1 + ∑ r = 1 49 1 2 r − ∑ r = 1 49 1 r − 1 100 \displaystyle \Rightarrow U = \sum _{ r=1 }^{ 50 }{ \frac { 1 }{ 2r-1 } } +\sum _{ r=1 }^{ 49 }{ \frac { 1 }{ 2r } } -\sum _{ r=1 }^{ 49 }{ \frac { 1 }{ r } } -\frac { 1 }{ 100 }

⇒ U = ∑ r = 1 99 1 r − ∑ r = 1 49 1 r − 1 100 \Rightarrow \displaystyle U = \sum _{ r=1 }^{ 99 }{ \frac { 1 }{ r } } -\sum _{ r=1 }^{ 49 }{ \frac { 1 }{ r } } -\frac { 1 }{ 100 }

⇒ U = ∑ r = 50 99 1 r − 1 100 = T − 1 100 \Rightarrow \displaystyle U = \sum _{ r=50 }^{ 99 }{ \frac { 1 }{ r } } -\frac { 1 }{ 100 } =T-\frac { 1 }{ 100 }

⇒ U = T − 1 100 \Rightarrow U = T - \dfrac{1}{100}

⇒ S = 1 100 \Rightarrow S = \dfrac{1}{100}

14 Helpful 0 Interesting 0 Brilliant 0 Confused

same way(+1).

Aareyan Manzoor - 5 years, 5 months ago

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