已知 {an}\left\{ {a}_{n}\right\}{an} 是各项均为正数的等差数列,公差为 ddd ,对任意的 n∈N∗, bnn \in N * ,{\mathrm{\;b}}_{n}n∈N∗,bn 是 an{a}_{n}an 和 an+1{a}_{n + 1}an+1 的等比中项.(1) 设 cn=bn+12−bn2,n∈N∗{c}_{n} = {b}_{n + 1}^{2} - {b}_{n}^{2},n \in {\mathrm{N}}^{ * }cn=bn+12−bn2,n∈N∗ ,求证: {cn}\left\{ {c}_{n}\right\}{cn} 是等差数列;(2) 设 a1=d,Tn=∑k=12n(−1)kbk2,n∈N∗{a}_{1} = d,{T}_{n} = \mathop{\sum }\limits_{{k = 1}}^{{2n}}{\left( -1\right) }^{k}{b}_{k}{}^{2},n \in {N}^{ * }a1=d,Tn=k=1∑2n(−1)kbk2,n∈N∗ ,求证: ∑k=1n1Tk<12d2\mathop{\sum }\limits_{{k = 1}}^{n}\frac{1}{{T}_{k}} < \frac{1}{2{d}^{2}}k=1∑nTk1<2d21 .