Prove that 1 2 ( x 2 )+1 4 ( x 4 )+...+1 2^ n ( x 2^ n )=1 2^ n ( x 2^ n )- for all n and 0<x< 2

Let p(n): 1 2 ( x 2 )+1 4 ( x 4 )+...+1 2^ n ( x 2^ n )=1 2^ n ( x 2^ n )- For n = 1 1 2 ( x 2 )=1 2 ( x 2 )- =1 2 1 2 x 2 -1 =1 2 x 2 -1 ( 2 x 2 1-2 ^2 x 2 ) =1 2 x 2 - 1- ^2 x 2 2 x 2 = 1-1+ ^2 x 2 2 x 2 = ^2 x 2 2 x 2 =1 2 x 2 ⇒ p(n) is true for n = 1 Let p(n) is true for n = k, So 1 2 ( x 2 )+1 4 ( x 4 )+...+1 2^ k ( x 2^ k )=1 2^ k ( x 2^ k )- We have to show that, 1 2 ( x 2 )+1 4 ( x 4 )+...+1 2^ k ( x 2^ k )+1 2^ k+ 1 ( x 2^ k+1 )=1 2^ k+1 ( x 2^ k+1 )- Now, 1 2 x 2 +1 4 ( x 4 )+...+1 2^ k ( x 2^ k ) +1 2^ k+ 1 ( x 2^ k+1 ) =1 2^k ( x 2^k )- +1 2^ k+1 ( x 2^ k+1 ) =1 2^k ( x 2^k )- +1 2.2^ k 1 ( x 2^k .1 2 ) =1 2^k [1 ( x 2^k ) +1 2 . ( x 2^k ).1 2 ]- =1 2^k [ 1- ^2 ( x 2^k+1 ) 2 ( x 2^k+1 ) +1 2 ( x 2.2^k ) ]- =1 2^k [ 1- ^2 ( x 2^ k+1 )+ ^2 ( x 2^k+1 ) 2 ( x 2^k+1 ) ]- =1 2^ k+1 [1 ( x 2^ k+1 ) ]- =1 2^ k+1 ( x 2^ k+1 )- ⇒ P(n) is true for n = k + 1 ⇒ P(n) is true for n by PMI

Prove that 1 2 ( x 2 )+1 4 ( x 4 )+...+1 2^ n ( x 2^ n )=1 2^ n (…

Download App

Question

Prove that $\frac{1}{2}\tan\Big(\frac{\text{x}}{2}\Big)+\frac{1}{4}\tan\Big(\frac{\text{x}}{4}\Big)+...+\frac{1}{2^{\text{n}}}\tan\Big(\frac{\text{x}}{2^{\text{n}}}\Big)=\frac{1}{2^{\text{n}}}\cot\Big(\frac{\text{x}}{2^{\text{n}}}\Big)-\cot\text{x}$ for all $\text{n}\in\text{N}$ and $0<\text{x}<\frac{\pi}{2}$

✓

Answer

Let p(n): $\frac{1}{2}\tan\Big(\frac{\text{x}}{2}\Big)+\frac{1}{4}\tan\Big(\frac{\text{x}}{4}\Big)+...+\frac{1}{2^{\text{n}}}\tan\Big(\frac{\text{x}}{2^{\text{n}}}\Big)=\frac{1}{2^{\text{n}}}\cot\Big(\frac{\text{x}}{2^{\text{n}}}\Big)-\cot\text{x}$
For n = 1
$\frac{1}{2}\tan\Big(\frac{\text{x}}{2}\Big)=\frac{1}{2}\cot\Big(\frac{\text{x}}{2}\Big)-\cot\text{x}$
$=\frac{1}{2}\frac{1}{2\tan\frac{\text{x}}{2}}-\frac{1}{\tan\text{x}}$
$=\frac{1}{2\tan\frac{\text{x}}{2}}-\frac{1}{\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1-2\tan^2\frac{\text{x}}{2}}\Bigg)}$
$=\frac{1}{2\tan\frac{\text{x}}{2}}-\frac{1-\tan^2\frac{\text{x}}{2}}{2\tan\frac{\text{x}}{2}}$
$=\frac{1-1+\tan^2\frac{\text{x}}{2}}{2\tan\frac{\text{x}}{2}}$
$=\frac{\tan^2\frac{\text{x}}{2}}{2\tan\frac{\text{x}}{2}}$
$=\frac{1}{2}\tan\frac{\text{x}}{2}$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, So
$\frac{1}{2}\tan\Big(\frac{\text{x}}{2}\Big)+\frac{1}{4}\tan\Big(\frac{\text{x}}{4}\Big)+...+\frac{1}{2^{\text{k}}}\tan\Big(\frac{\text{x}}{2^{\text{k}}}\Big)=\frac{1}{2^{\text{k}}}\cot\Big(\frac{\text{x}}{2^{\text{k}}}\Big)-\cot\text{x}$
We have to show that,
$\frac{1}{2}\tan\Big(\frac{\text{x}}{2}\Big)+\frac{1}{4}\tan\Big(\frac{\text{x}}{4}\Big)+...+\frac{1}{2^{\text{k}}}\tan\Big(\frac{\text{x}}{2^{\text{k}}}\Big)+\frac{1}{2^{\text{k+ 1}}}\tan\Big(\frac{\text{x}}{2^{\text{k+1}}}\Big)=\frac{1}{2^{\text{k+1}}}\cot\Big(\frac{\text{x}}{2^{\text{k+1}}}\Big)-\cot\text{x}$
Now,
$\Bigg\{\frac{1}{2}\tan\frac{\text{x}}{2}+\frac{1}{4}\tan\Big(\frac{\text{x}}{4}\Big)+...+\frac{1}{2^{\text{k}}}\tan\Big(\frac{\text{x}}{2^{\text{k}}}\Big)\Bigg\}+\frac{1}{2^{\text{k+ 1}}}\tan\Big(\frac{\text{x}}{2^{\text{k+1}}}\Big)$
$=\frac{1}{2^\text{k}}\cot\Big(\frac{\text{x}}{2^\text{k}}\Big)-\cot\text{x}+\frac{1}{2^{\text{k+1}}}\tan\Big(\frac{\text{x}}{2^{\text{k}+1}}\Big)$
$=\frac{1}{2^\text{k}}\cot\Big(\frac{\text{x}}{2^\text{k}}\Big)-\cot\text{x}+\frac{1}{2.2^{\text{k}}}\frac{1}{\cot\Big(\frac{\text{x}}{2^\text{k}}.\frac{1}{2}\Big)}$
$=\frac{1}{2^\text{k}}\Bigg[\frac{1}{\tan\Big(\frac{\text{x}}{2^\text{k}}\Big)}+\frac{1}{2}.\tan\bigg\{\Big(\frac{\text{x}}{2^\text{k}}\Big).\frac{1}{2}\bigg\}\Bigg]-\cot\text{x}$
$=\frac{1}{2^\text{k}}\Bigg[\frac{1-\tan^2\Big(\frac{\text{x}}{2^\text{k+1}}\Big)}{2\tan\Big(\frac{\text{x}}{2^\text{k+1}}\Big)}+\frac{1}{2}\tan\Big(\frac{\text{x}}{2.2^\text{k}}\Big)\Bigg]-\cot\text{x} $
$=\frac{1}{2^\text{k}}\Bigg[\frac{1-\tan^2\Big(\frac{\text{x}}{2^{\text{k}+1}}\Big)+\tan^2\Big(\frac{\text{x}}{2^\text{k+1}}\Big)}{2\tan\Big(\frac{\text{x}}{2^\text{k+1}}\Big)}\Bigg]-\cot\text{x}$
$=\frac{1}{2^{\text{k}+1}}\Bigg[\frac{1}{\tan\Big(\frac{\text{x}}{2^{\text{k}+1}}\Big)}\Bigg]-\cot\text{x}$
$=\frac{1}{2^{\text{k}+1}}\cot\Big(\frac{\text{x}}{2^{\text{k}+1}}\Big)-\cot\text{x}$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for $\text{n}\in\text{N}$ by PMI

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Mathematical Induction→Mathematical Induction — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Mathematical Induction — Maths STD 11 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions