## Solving Lesson 3 Page 37 Math Learning Topic 10 – Kite>

Topic

Prove $$C_n^0{3^n} + C_n^1{3^{n – 1}} + … + C_n^k{3^{n – k}} + … + C_n^{n – 1} 3 + C_n^n$$

$$= C_n^03 + C_n^13 + … + C_n^k{3^k} + … + C_n^{n – 1}{3^{n – 1}} + C_n^n{.3^n}$$ with $$0 \le k \le n,n \in \mathbb{N}$$

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

Applying Newton’s binomial formula we have:

$${\left( {a + b} \right)^n} = C_n^0. {a^n}. {b^0} + C_n^1{a^{n – 1}}. {b^ 1} + … + C_n^k{a^{n – k}} {b^k} + … + C_n^{n – 1}a {b^{n – 1}} + C_n^n. { a^0}. {b^n}$$

Replace $$a = 3,b = 1$$ we get

$$\begin{array}{l} \Leftrightarrow {\left( {3 + 1} \right)^n} = C_n^0{.3^n}{.1^0} + C_n^1{3^ {n – 1}}{.1^1} + … + C_n^k{3^{n – k}}{.1^k} + … + C_n^{n – 1}{3.1^{n – 1 }} + C_n^n{.3^0}{.1^n}\\ \Rightarrow {4^n} = C_n^0{3^n} + C_n^1{3^{n – 1}} + … + C_n^k{3^{n – k}} + … + C_n^{n – 1}3 + C_n^n\end{array}$$

Replace $$a = 1,b = 3$$ we get

$$\begin{array}{l}{\left( {1 + 3} \right)^n} = C_n^0{.1^n}{.3^0} + C_n^1{1^{n – 1}}{.3^1} + … + C_n^k{1^{n – k}}{.3^k} + … + C_n^{n – 1}{1.3^{n – 1}} + C_n^n{.1^0}{.3^n}\\ \Rightarrow {4^n} = C_n^03 + C_n^13 + … + C_n^k{3^k} + … + C_n^{ n – 1}{3^{n – 1}} + C_n^n{.3^n}\end{array}$$

Figure out what to prove