KloppsAndRobbers
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Been trying for ages - atp I just want to see it solved. Please help me lol.
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Mathematic Induction Question (1 Viewer)
- Thread starter KloppsAndRobbers
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member 6003
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Assume for n=k
\theta}=\tan{(k+1)\theta}\cot{\theta}-(k+1) )
To prove for n=k+1
That is
\theta}\tan{(k+2)\theta}=\tan{(k+2)\theta}\cot{\theta}-(k+1))
\theta}+\tan{(k+1)\theta}\tan{(k+2)\theta})
using assumption,
\theta}\cot{\theta}-(k+1)+ \tan{(k+1)\theta}\tan{(k+2)\theta} )
 + 1 + \tan{(k+1)\theta}\cot{\theta} + \tan{(k+1)\theta}\tan{(k+2)\theta} )
 + \cot{\theta}[ \tan{\theta} + \tan{(k+1)\theta} + \tan{(k+1)\theta}\tan{(k+2)\theta}\tan{\theta} ])
using i) for\theta)
 + \cot{\theta}[(\tan{(k+2)\theta})(1-\tan{\theta}\tan{(k+1)\theta}) + \tan{(k+1)\theta}\tan{(k+2)\theta}\tan{\theta}])
 + \cot{\theta}[\tan{(k+2)\theta} - \tan{\theta}\tan{(k+1)\theta}\tan{(k+2)\theta} + \tan{(k+1)\theta}\tan{(k+2)\theta}\tan{\theta}])
\theta}\cot{\theta} - (k+2))
therefore result follows by induction
sorry for bracket mistakes, this question you just want try to match very closely to the RHS or your not gonna get it because of the +1 term
To prove for n=k+1
That is
using assumption,
using i) for
therefore result follows by induction
sorry for bracket mistakes, this question you just want try to match very closely to the RHS or your not gonna get it because of the +1 term
Last edited:
KloppsAndRobbers
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- 2024
Thank you so much 