Đáp án:
` t=2^{2020}`
`s=2^{14}`
Giải thích các bước giải:
Ta có: ${(1 + x)^{2021}} = C_{2021}^0 + C_{2021}^1x + C_{2021}^2{x^2} + ... + C_{2021}^{2020}{x^{2020}}C_{2021}^{2021}{x^{2021}}$
Với `x=1 =>` ${2^{2021}} = C_{2021}^0 + C_{2021}^1 + ... + C_{2021}^{2021}$ `(1)`
Với `x=-1 =>` $0 = C_{2021}^0 - C_{2021}^1 + ... + C_{2021}^{2020} - C_{2021}^{2021}$ `(2)`
`(1)+(2)` $ = 2(C_{2021}^0 + C_{2021}^2 + ... + C_{2021}^{2020}) = {2^{2021}}$
`=>` $C_{2021}^0 + C_{2021}^2 + ... + C_{2021}^{2020} = {2^{2020}}$
Vậy `t=2^{2020}`
Ta có: ${(1 + x)^{15}} = C_{15}^0 + C_{15}^1x + .... + C_{15}^{15}{x^{15}}$
Với `x=1 =>` $C_{15}^0 + C_{15}^1 + .... + C_{15}^{15} = {2^{15}}$
mà $\left\{ \begin{array}{l} C_{15}^0 = C_{15}^{15}\\ C_{15}^1 = C_{15}^{14}\\ ...\\ C_{15}^7 = C_{15}^8 \end{array} \right. \Rightarrow 2(C_{15}^8 + C_{15}^9 + ... + C_{15}^{15}) = {2^{15}}$
`=>` $C_{15}^8 + C_{15}^9 + ... + C_{15}^{15} = {2^{14}}$
Vậy `s=2^{14}`
