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I need to prove the equation $$ \frac 1 3 = \frac { 1 + 3 + 5 + \dots + ( 2 n - 1 ) } { ( 2 n + 1 ) + ( 2 n + 3 ) + \dots + \bigl ( 2 n + ( 2 n - 1 ) \bigr ) } $$ using mathematical induction.

I tried solving this but I got stuck. I would be very thankful is someone could help me. Or maybe give me a references or hint.

My Attempt:

  • For $n=1$

$$\frac13=\frac1{2n+1}$$

$$\frac13=\frac1{2+1}$$

$$\frac13=\frac1{3}$$

which is true.

  • For $n=k$

$$ \frac 1 3 = \frac { 1 + 3 + 5 + \dots + ( 2 k - 1 ) } { ( 2 k + 1 ) + ( 2 k + 3 ) + \dots + \bigl ( 2 k + ( 2 k - 1 ) \bigr ) } $$

  • For $n=k+1$

$$ \frac 1 3 = \frac { 1 + 3 + 5 + \dots + ( 2 k - 1 )+ ( 2 k+1 - 1 ) } { ( 2 k + 1 ) + ( 2 k + 3 ) + \dots + \bigl ( 2 k + ( 2 k - 1 ) \bigr ) } $$

user
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user123awq
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1 Answers1

1

You did some mistake in the expression for the case $n=k+1$ in the induction step. Morover it is convenient proceed as follows using that $\frac A B=\frac13 \iff 3A=B$.

We need to show by induction that that for any $n\ge 1$

$$\frac13 = \sum_{i=1}^n \frac { 1 + 3 + 5 + \dots + ( 2 i - 1 ) } { ( 2 i + 1 ) + ( 2 i + 3 ) + \dots + \bigl ( 2 i + ( 2 i - 1 ) \bigr ) }$$

  • base case: $n=1 \implies \frac13 = \frac1{2+1}$
  • induction step we assume

$$\frac13 = \sum_{i=1}^n \frac { 1 + 3 + 5 + \dots + ( 2 i - 1 ) } { ( 2 i + 1 ) + ( 2 i + 3 ) + \dots + \bigl ( 2 i + ( 2 i - 1 ) \bigr ) }$$

$$\iff 3\sum_{i=1}^n\left(1 + 3 + 5 + \dots + ( 2 i - 1 )\right)=\sum_{i=1}^n( 2 i + 1 ) + ( 2 i + 3 ) + \dots + \bigl ( 2 i + ( 2 i - 1 ) \bigr )$$

then

$$ 3\sum_{i=1}^{n+1}\left(1 + 3 + 5 + \dots + ( 2 i - 1 )\right)=\sum_{i=1}^{n+1}( 2 i + 1 ) + ( 2 i + 3 ) + \dots + \bigl ( 2 i + ( 2 i - 1 ) \bigr )$$

$$\iff 3(2(n+1)-1)=2n+2(n+1)+(2(n+1)-1)$$

$$\iff 3(2n+1)=2n+2n+2+(2n+1)$$

$$\iff 6n+3=4n+2+(2n+1)$$

$$\iff 6n+3=6n+3$$

user
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