Cho dãy số un thỏa mãn u1=1;un+1= căn au^2n+1

* Hướng dẫn giải

Đáp án A

Ta có $u_{n+1}=\sqrt{a{u}_n^2+1}\iff u_{n+1}^2=a{u}_n^2+1\iff u_{n+1}^2-\frac{1}{1-a}=a\left(u_n^2-\frac{1}{1-a}\right)$ .

Đặt $v_n=u_n^2-\frac{1}{1-a}\Rightarrow v_{n+1}=a{v}_n\Rightarrow \left(v_n\right)$  là cấp số nhân với công bội q = 0.

Suy ra $v_n=v_1{a}^{n-1}=\left(u_1^2-\frac{1}{1-a}\right)a^{n-1}=a^{n-1}.\frac{a}{a-1}\Rightarrow u_n^2=a^{n-1}.\frac{a}{a-1}+\frac{1}{1-a}$ .

Ta có:$\left\{\begin{array}{l}{u}_1^2=\frac{a}{a-1}+\frac{1}{1-a}\\ u_2^2=a.\frac{a}{a-1}+\frac{1}{1-a}\\ \mathrm{.............................}\\ u_n^2=a^{n-1}.\frac{a}{a-1}+\frac{1}{1-a}\end{array}\right.\Rightarrow u_1^2+u_2^2+\mathrm{...}+u_n^2=\frac{a}{a-1}\left(1+a+\mathrm{...}+a^{n-1}\right)+\frac{1}{1-a}.n$

. Khi đó$\left\{\begin{array}{l}\frac{1}{1-a}=2\Rightarrow a=\frac{1}{2}\\ b=lim\left(\frac{a}{a-1}.\frac{1-a^n}{1-a}\right)=-2\end{array}\right.\Rightarrow T=-1$ .