Giải Hoạt động 3 trang 50 Toán 11 tập 1 Cánh diều

a) Ta có: $u_{1} + u_{n} = u_{1} + u_{1} + (n – 1)d = 2u_{1} + (n – 1)d$;

$u_{2} + u_{n-1}­ = u_{1} + d + u_{1} + (n – 1 – 1)d = 2u_{1} + (n – 1)d$;

$u_{3} + u_{n-2} = u_{1} + 2d + u_{1} + (n – 2 – 1)d = 2u_{1} + (n – 1)d$;

...

$u_{n} + u_{1} = u_{1} + (n – 1)d + u_{1} = 2u_{1} + (n – 1)d.$

Ta thấy $u_{1} + u_{n} = u_{2} + u_{n-1} = u_{3} + u_{n-2} = ... = u_{n} + u_{1}$.

b) Ta có: $2S_{n} = 2.(u_{1} + u_{2} + u_{3} + ... + u_{n}) = (u_{1} + u_{n}) + (u_{2} + u_{n-1}) + ... + (u_{n} + u_{1})$

= $2u_{1} + (n – 1)d + 2u_{1} + (n – 1)d + 2u_{1} + (n – 1)d + ... + 2u_{1} + (n – 1)d$

= $2n.u_{1} + n(n – 1)d$

= $n(u_{1} + u_{1} + (n – 1)d)$

= $n(u_{1} + u_{n}).$