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a
(1+x)n>(1+nx) for all natural numbers n
b
(1+x)n≥(1+nx) for all natural numbers z, where x > -1
c
(1+x)n≤(1+nx) for all natural numbers n
d
(1+x)n<(1+nx) for all natural numbers n
answer is B.
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Detailed Solution
Let P(n): (1+x)n≥(1+nx)For n=1, (1+x)1=1+x=1+1⋅x≥1+1⋅x (1+x)1≥1+1⋅xFor n=k, let P(k):(1+x)k≥(1+kx) is true. For n=k+1,P(k+1):(1+x)k+1≥{1+(k+1)x} we shall show P (k + 1) is true. consider,(1+x)k+1=(1+x)k⋅(1+x)≥(1+kx)(1+x) [ if x>−1]=1+x+kx+kx2≥1+x+kx [∵k>0 and x>−1]=1+(k+1)xThus,(1+x)k+1≥1+(k+1)x, if x>−1