已知数列 {an}\left\{ {a}_{n}\right\}{an} 满足 a1=1{a}_{1} = 1a1=1 ,且对任意正整数 m,nm,nm,n 都有 am+n=an+am+2mn.{a}_{m + n} = {a}_{n} + {a}_{m} + {2mn}.am+n=an+am+2mn.(1)求数列 {an}\left\{ {a}_{n}\right\}{an} 的通项公式;(2)求数列 {(−1)nan}\left\{ {{\left( -1\right) }^{n}{a}_{n}}\right\}{(−1)nan} 的前 nnn 项和 Sn{S}_{n}Sn ,若存在正整数 kkk ,使得 2Sk+Sk+1=02{S}_{k} + {S}_{k + 1} = 02Sk+Sk+1=0 ,求 kkk 的值;(3) 设 bn=12ln(1+1an+1an+1),Tn{b}_{n} = \frac{1}{2}\ln \left( {1 + \frac{1}{{a}_{n}} + \frac{1}{{a}_{n + 1}}}\right) ,{T}_{n}bn=21ln(1+an1+an+11),Tn 是数列 {bn}\left\{ {b}_{n}\right\}{bn} 的前 nnn 项和,求证: Tn<nn+1{T}_{n} < \frac{n}{n + 1}Tn<n+1n .