Diberikan P(n) : 11+nx≥1(1+x)nP\left(n\right)\ :\ \frac{1}{1+nx}\geq \frac{1}{(1+x)^n}P(n) : 1+nx1≥(1+x)n1 untuk setiap bilangan asli nnn dan suatu bilangan bulat xxx. Jika P(n)P\left(n\right)P(n) benar untuk n=k+1n=k+1n=k+1, maka berlaku ....
11+kx≥1(1+x)k\frac{1}{1+kx}\geq \frac{1}{(1+x)^k}1+kx1≥(1+x)k1
11+x+kx≥1(1+x)k\frac{1}{1+x+kx}\geq \frac{1}{(1+x)^k}1+x+kx1≥(1+x)k1
11+x+kx≥1(1+x)k+1\frac{1}{1+x+kx}\geq \frac{1}{(1+x)^{k+1}}1+x+kx1≥(1+x)k+11
11+kx≥1(1+x)k+1\frac{1}{1+kx}\geq \frac{1}{(1+x)^{k+1}}1+kx1≥(1+x)k+11
11+k+kx≥1(1+x)k+1\frac{1}{1+k+kx}\geq \frac{1}{(1+x)^{k+1}}1+k+kx1≥(1+x)k+11