Question

решить методом математической индукции.. только 12.45​

Answer 1

$frac{1}{5cdot12}+frac{1}{12cdot19}+...+frac{1}{(7n-2)(7n+5)}=frac{n}{5(7n+5)}$

1.

$n=1$

$frac{1}{5cdot12}=frac{1}{5cdot(7cdot1+5)}$

$frac{1}{5cdot12}=frac{1}{5cdot(7+5)}$

$frac{1}{5cdot12}=frac{1}{5cdot12}$

2.

$n=k$

$frac{1}{5cdot12}+frac{1}{12cdot19}+...+frac{1}{(7k-2)(7k+5)}=frac{k}{5(7k+5)}$

3.

$n=k+1$

$frac{1}{5cdot12}+frac{1}{12cdot19}+...+frac{1}{(7(k+1)-2)(7(k+1)+5)}=frac{1}{5cdot12}+frac{1}{12cdot19}+...+frac{1}{(7(k+1)-2)(7(k+1)+5)}=frac{k+1}{5(7(k+1)+5)}$

$frac{k}{5(7k+5)}+frac{1}{(7(k+1)-2)(7(k+1)+5)}=$

$frac{k}{5(7k+5)}+frac{1}{(7k+7-2)(7k+7+5)}=$

$frac{k}{5(7k+5)}+frac{1}{(7k+5)(7k+12)}=$

$frac{k(7k+12)}{5(7k+5)}+frac{5}{5(7k+5)(7k+12)}=$

$frac{7k^2+12k+5}{5(7k+5)(7k+12)}=$

$frac{7k^2+5k+7k+5}{5(7k+5)(7k+12)}=$

$frac{k(7k+5)+(7k+5)}{5(7k+5)(7k+12)}=$

$frac{(7k+5)(k+1)}{5(7k+5)(7k+12)}=$

$frac{k+1}{5(7k+12)}=$

$frac{k+1}{5(7k+7+5)}=$

$frac{k+1}{5(7(k+1)+5)}$

Answer 2

помогите

Answer 3

пожалуйста последний вопрос