Binomial Theorem And Its Simple Applications Question 206
Question: The greatest integer less than or equal to; $ {{(\sqrt{2}+1)}^{6}} $ is
Options:
A) 196
B) 197
C) 198
D) 199
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Let $ {{(\sqrt{2}+1)}^{6}}=k+f, $ where k is integral part and f the fraction $ (0\le f<1) $ Let $ {{(\sqrt{2}-1)}^{6}}=f’,(0<f’<1), $ Since $ 0<(\sqrt{2}-1)<1 $ Now, $ k+f+f’={{(\sqrt{2}+1)}^{6}}+{{(\sqrt{2}-1)}^{6}} $
$ \therefore f+f’=198-k $ = an integer But $ 0\le f<1 $ and $ 0<f’<1\Rightarrow 0<(f+f’)<2 $
$ \Rightarrow f+f’=1,(\because f+f’isaninteger) $
$ \therefore By(i),I=198-(f+f’)=198-1=197 $
BETA
