Question

решить методом математической индукции.. только 12.45​

Answer 1

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

1.

$n=1$

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

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

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

2.

$n=k$

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

3.

$n=k+1$

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

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

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

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

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

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

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

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

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

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

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

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