## Solving Lesson 2 Page 37 Math Learning Topic 10 – Kite>

Topic

Calculate:

a) $$S = C_{2022}^0{9^{2022}} + C_{2022}^1{9^{2021}} + … + C_{2022}^k{9^{2022 – k} } + … + C_{2022}^{2021}9 + C_{2022}^{2022}$$

b) $$T = C_{2022}^0{4^{2022}} – C_{2022}^1{4^{2021}}.3 + … – C_{2022}^{2021}{4.3^{ 2021}} + C_{2022}^{2022}{.3^{2022}}$$

Solution method – See details

Newton’s binomial formula: $${(a + b)^n} = C_n^0{a^n} + C_n^1{a^{n – 1}}b + … + C_n^{n – 1} a{b^{n – 1}} + C_n^n{b^n}$$

Detailed explanation

a) According to Newton’s binomial formula, we have: $${\left( {9 + x} \right)^{2022}} = C_{2022}^0{9^{2022}}. {x^0 } + C_{2022}^1{9^{2021}} {x^1} + … + C_{2022}^k{9^{2022 – k}} {x^k} + … + C_{ 2022}^{2021}9. {x^{2021}} + C_{2022}^{2022}. {x^{2022}}$$

Replace $$x = 1$$ we get: $${\left( {9 + 1} \right)^{2022}} = S = C_{2022}^0{9^{2022}} + C_{2022 }^1{9^{2021}} + … + C_{2022}^k{9^{2022 – k}} + … + C_{2022}^{2021}9 + C_{2022}^{2022} \ Rightarrow S = {10^{2022}}$$

b) According to Newton’s binomial formula, we have:

$${\left( {4 + x} \right)^{2022}} = C_{2022}^0{4^{2022}} {x^0} + C_{2022}^1{4^{ 2021}} {x^1} + … + C_{2022}^k{4^{2022 – k}} {x^k} + … + C_{2022}^{2021}4. {x^{ 2021}} + C_{2022}^{2022}. {x^{2022}}$$

Replace $$x = – 3$$ we get

$$\begin{array}{l}{\left( {4 – 3} \right)^{2022}} = C_{2022}^0{4^{2022}}. {\left( { – 3} \right)^0} + C_{2022}^1{4^{2021}} {\left( { – 3} \right)^1} + …… + C_{2022}^{2021}4. { \left( { – 3} \right)^{2021}} + C_{2022}^{2022}. {\left( { – 3} \right)^{2022}}\\ \Leftrightarrow {1^{2022 }} = T = C_{2022}^0{4^{2022}} – C_{2022}^1{4^{2021}}.3 + … – C_{2022}^{2021}{4.3^{2021 }} + C_{2022}^{2022}{.3^{2022}}\\ \Leftrightarrow T = 1\end{array}$$