每日一题_191219

若数列({a_n})满足(a_{n+1}=3a_n+2),(ninmathbb{N}^ast),且(a_1=2).
((1)) 求数列({a_n})的通项公式;
((2)) 记(b_n=dfrac{1}{a_n^2}+dfrac{1}{a_n}),数列({b_n})的前(n)项和为(S_n),求证:(forall ninmathbb{N}^ast,S_n<dfrac{31}{32}).

解析:
((1)) 由题$$
a_{n+1}+1=3left(a_n+1 ight).$$所以$$
a_n=3^{n-1,ninmathbb{N}}ast.$$
((2)) 结合((1))可知(b_n=dfrac{3^n}{left(3^n-1 ight)^2}),所以$$
forall ngeqslant 2,b_n=dfrac{3^{{n-1}}{left(3}n-1 ight)left(3^{{n-1}-dfrac{1}{3} ight)}<dfrac{1}{2}left(dfrac{1}{3}{n-1}-1}-dfrac{1}{3^n-1} ight).$$
从而
$$
egin{split}
S_n&leqslant b_1+b_2+sum_{k=3}^{n}left[dfrac{1}{2}left( dfrac{1}{3^{{k-1}-1}-dfrac{1}{3}k-1} ight) ight]
&=dfrac{3}{4}+dfrac{9}{64}+dfrac{1}{16}-dfrac{1}{2}cdotdfrac{1}{3^n-1}
&<dfrac{61}{64}<dfrac{31}{32}.
end{split}
$$
证毕.