qqq makaron: spr Wykaż,że dla dowolnej liczby n>0 i n≠1 log_n(n+1)*log_n+1(n+2)*log_n+2(n+3)*log_n+3n = 1 robię tak

	logn(n+2)		logn+2n
logn(n+1)*		*logn+2(n+3)*		=
	logn(n+1)		logn+2(n+3)

	lognn
logn(n+2)*		= 1
	logn(n+2)

log_nn = 1 1 = 1 cnd ?