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![Sample Induction Proofs]()
www.math.uiuc.eduInduction step: Let k ∈ Z+ be given and suppose (1) is true for n = k. Then .... Prove that for any real number x > −1 and any positive integer x, (1 + x)n ≥ 1 + nx .瀏覽：470 -
![Proof that (1+x)^n >= 1+nx? - Straight Dope Message Board]()
boards.straightdope.comProve that (1+x)^n >= 1+nx for x>-1 and n any natural number. ... -1) to get (1+x)n +1 >= ... something which can be shown to be >= 1 + (n+1)x.瀏覽：943
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瀏覽：525
日期：2024-07-20
Prove that (1+x)^n >= 1+nx for x>-1 and n any natural number. ... -1) to get (1+x)n
+1 >= ... something which can be shown to be >= 1 + (n+1)x....
瀏覽：743
日期：2024-07-22
Prove by induction that (1 + x)^n >= (1 + nx), where n is a non- negative integer.
... (1+x)^(k+1) >= (1 + (k+1)x) >= (1 + kx + x) >= (1 + kx) + x So ......
瀏覽：1246
日期：2024-07-17
Given: (d/dx) e = e; (d/dx) ln(x) = 1/x; Chain Rule. Solve: (d/dx) xn = (d/dx) e(n ln x
). = (d/du) eu (d/dx) (n ln x) (Set u = n ln x). = [e(n ln x)] [n/x] = xn n/x = n x(n-1) ......
瀏覽：919
日期：2024-07-23
e = e; (d/dx) ln(x) = 1/x; Chain Rule. Solve: (d/dx) x^n = (d/dx) e^(n ln x). = (d/du) e
^u (d/dx) (n ln x) (Set u = n ln x). = [e^(n ln x)] [n/x] = x^n n/x = n x^(n-1) ......
瀏覽：420
日期：2024-07-19
(1+x)^2 = (1+2x+x^2) > 1+2x. Assume (1+x)^n > 1+nx, then (1+x)^(n+1) = (1+x) (1
+x)^n > (1+x)(1+nx) = 1+ (n+1)x + nx^2 > 1+ (n+1)x. 0. 0....
瀏覽：737
日期：2024-07-23
19 Apr 2014 ... firstly, we can know it is true for n=1 and n=2,suppose when n=k, it is also true. it
means that >=1+kx ... Finally,(1+kx)(1+x)=1+kx+x+ >=1+(k+1)x....
瀏覽：1106
日期：2024-07-21
Step 1: Proof of the Power Rule for Non-Negative Integer Exponents. In this step
... f ′(x) = xn − 1 + xn − 1 + ... + xn − 1. There are. n. terms, so. f ′(x) = nxn − 1,....
瀏覽：353
日期：2024-07-23
1 1 1 2 1 1 3 3 1. If you continued expanding the brackets for higher powers, you
would find that ... (1 + x)n, = 1, +, nx, +, n(n - 1)x2, +, n(n - 1)(n - 2)x3, + ... 1! 2! 3!...
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