A) \[{{(1+x)}^{n}}>(1+nx),\] for all natural numbers \[n\]
B) \[{{(1+x)}^{n}}\ge (1+nx),\]For natural numbers \[n\]where \[x>-1\]
C) \[{{(1+x)}^{n}}\le (1+nx),\]for all natural numbers \[n\]
D) \[{{(1+x)}^{n}}<(1+nx),\]for all natural numbers \[n\]
Correct Answer: B
Solution :
| Let \[\operatorname{P}(n):{{(1+x)}^{n}}\ge (1+nx)\] |
| For \[n=1,\,{{(1+x)}^{1}}=1+x\]\[=1+{{1}^{.}}x\ge 1+{{1}^{.}}x{{(1+x)}^{1}}\ge 1+{{1}^{.}}x\] |
| For \[n=k,+1,P(k):{{(1+x)}^{k}}\ge (1+kx)\]is true. |
| For \[n=k+1,P(k+1):{{(1+x)}^{k+1}}\]\[\ge \{1+(k+1)x\}\]is also true. |
| We will show \[P(k+1)\]is true. |
| Consider\[{{(1+x)}^{k+1}}={{(1+x)}^{k.}}(1+x)\]\[\ge (1+kx)(1+x)[if\,x>-1]\] |
| \[=1+x+kx+k{{x}^{2}}\ge 1+x+kx\] |
| \[\left[ \because k>0\operatorname{and}x>-1 \right]\] |
| \[=1+(k+1)x\] |
| Thus, \[{{(1+x)}^{k+1}}\ge 1+(k+1)x,\,\operatorname{ifx}>-1\] |
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