広島大学先進理工系科学研究科情報科学プログラム 2018年1月実施専門科目I 問題5

标签：

Author

祭音Myyura

(1) $(a+b)^5$ を展開したとき、 $a^3b^2$ の項の係数を求めよ。

(2) 以下の等式の $X_1$ , $X_2$ , $X_3$ の各係数値を求めよ。

(a+b)^n = X_0 a^n + X_1 a^{n-1}b + X_2 a^{n-2}b^2 + X_3 a^{n-3}b^3 + \cdots + X_{n-1}ab^{n-1} + X_nb^n

(3) $11^5 = (10 + 1)^5$ を計算せよ。

(4) $(a + b + c)^7$ を展開したとき、 $a^2 b^3 c^2$ の項の係数を求めよ。

(5) $(2a + b + c)^7$ を展開したとき、 $a^2 b^3 c^2$ の項の係数を求めよ。

(1) Consider the expansion of $(a+b)^5$ . Answer the coefficient of $a^3b^2$ .

(2) Answer each value of the coefficients $X_1$ , $X_2$ , $X_3$ of the following equation.

(a+b)^n = X_0 a^n + X_1 a^{n-1}b + X_2 a^{n-2}b^2 + X_3 a^{n-3}b^3 + \cdots + X_{n-1}ab^{n-1} + X_nb^n

(3) Calculate $11^5 = (10 + 1)^5$ .

(4) Consider the expansion of $(a + b + c)^7$ . Answer the coefficient of $a^2 b^3 c^2$ .

(5) Consider the expansion of $(2a + b + c)^7$ . nswer the coefficient of $a^2 b^3 c^2$ .

(a+b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5

Hence the coefficient of $a^3b^2$ is $10$ .

(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}

Hence $X_1 = n$ , $X_2 = \frac{n(n-1)}{2}$ , $X_3 = \frac{n(n-1)(n-2)}{6}$ .

\begin{aligned} 11^5 &= (10 + 1)^5 \\ &= 10^5 + 5 \times 10^4 + 10 \times 10^3 + 10 \times 10^2 + 5 \times 10 + 1\\ &= 100000 + 50000 + 10000 + 1000 + 50 + 1 \\ &= 161051 \end{aligned}

\begin{aligned} (a+b+c)^7 &= ((a+b) + c)^7 \\ &= (a+b)^7 + 7 (a+b)^6 c + 21 (a+b)^5 c^2 + \cdots \end{aligned}

\begin{aligned} 21 (a+b)^5 c^2 = 21(a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5)c^2 \end{aligned}

Hence the coefficient of $a^2b^3c^2$ is $21 \times 10 = 210$ .

\begin{aligned} (2a+b+c)^7 &= ((2a+b) + c)^7 \\ &= (2a+b)^7 + 7 (2a+b)^6 c + 21 (2a+b)^5 c^2 + \cdots \end{aligned}

\begin{aligned} 21 (2a+b)^5 c^2 = 21(2^5a^5 + 5 \times 2^4a^4 b + 10 \times 2^3 a^3 b^2 + 10 \times 2^2 a^2 b^3 + 5 \times 2 a b^4 + b^5)c^2 \end{aligned}

Hence the coefficient of $a^2b^3c^2$ is $21 \times 10 \times 2^2 = 840$ .