Why is the line in bold valid (Picture added below)?

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Answer 1 · 2017-07-22T17:34:29

Refer to the Explanation.

So far, it has been proved, that,

$(1+p)^(k+1) ge 1+kp+p+kp^2....(ast^1).$

Now, note that, $p > -1 rArr p^2 > 0." Also, "k in ZZ^+.$

$:. kp^2 > 0.$

Adding, on both sides of this inequality $1+kp+p,$ we get,

$kp^2+1+kp+p > 1+kp+p.............(ast^2).$

Combining, $(ast^1) and (ast^2),$ we have,

$(1+p)^(k+1) >= 1+kp+p+p^2 > 1+kp+p, i.e.,$

$(1+p)^(k+1) > 1+(k+1)p.$

In simple form, the line in bold means that,

if, $a >= b+c, &, c >0,$ then, if we drop $c >0,$ from $b+c,$

we still have, $a > c.$