Khai triển nhị thức (Px): {x^2} + ( 1/2) ^ 5 ta được

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Khai triển nhị thức \(P\left( x \right) = {\left( {{x^2} + \frac{1}{2}} \right)^5}\) ta được

\[C_5^0.{x^{10}} - C_5^1.{x^8}.\frac{1}{2} + C_5^2.{x^6}.{\left( {\frac{1}{2}} \right)^2} - C_5^3.{x^4}.{\left( {\frac{1}{2}} \right)^3} + C_5^4.{x^2}.{\left( {\frac{1}{2}} \right)^4} - C_5^5.{\left( {\frac{1}{2}} \right)^5}\].

\[C_5^0.{x^{10}} + C_5^1.{x^8}.2 + C_5^2.{x^6}{.2^2} + C_5^3.{x^4}{.2^3} + C_5^4.{x^2}{.2^4} + C_5^5{.2^5}\].

\[C_5^0.{x^{10}} + C_5^1.{x^8}.\frac{1}{2} + C_5^2.{x^6}.{\left( {\frac{1}{2}} \right)^2} + C_5^3.{x^4}.{\left( {\frac{1}{2}} \right)^3} + C_5^4.{x^2}.{\left( {\frac{1}{2}} \right)^4} + C_5^5.{\left( {\frac{1}{2}} \right)^5}\].

\[C_5^0.{x^{10}} - C_5^1.{x^8}.2 + C_5^2.{x^6}{.2^2} - C_5^3.{x^4}{.2^3} + C_5^4.{x^2}{.2^4} - C_5^5{.2^5}\].

Giải thích

Xét khai triển nhị thức \[P\left( x \right) = {\left( {{x^2} + \frac{1}{2}} \right)^5} = \sum\limits_{k = 0}^5 {C_5^k{{\left( {{x^2}} \right)}^{5 - k}}{{\left( {\frac{1}{2}} \right)}^k}} = \sum\limits_{k = 0}^5 {C_5^k{x^{10 - k}}{{\left( {\frac{1}{2}} \right)}^k}} \]

\[ = C_5^0.{x^{10}} + C_5^1.{x^8}.\frac{1}{2} + C_5^2.{x^6}.{\left( {\frac{1}{2}} \right)^2} + C_5^3.{x^4}.{\left( {\frac{1}{2}} \right)^3} + C_5^4.{x^2}.{\left( {\frac{1}{2}} \right)^4} + C_5^5.{\left( {\frac{1}{2}} \right)^5}\].