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
Author
祭音Myyura
Description
(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 \(a^2 b^3 c^2\).
Kai
(1)
\[
(a+b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5
\]
Hence the coefficient of \(a^3b^2\) is \(10\).
(2)
(Binomial coefficient)
\[
(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}
\]
Hence \(X_1 = n\), \(X_2 = \frac{n(n-1)}{2}\), \(X_3 = \frac{n(n-1)(n-2)}{6}\).
(3)
\[
\begin{aligned}
11^5 &= (10 + 1)^5 \\
&= 10^5 + 5 \times 10^4 + 10 \times 10^3 + 10 \times 10^2 + 5 \times 10 + 1\\
&= 100000 + 50000 + 10000 + 1000 + 50 + 1 \\
&= 161051
\end{aligned}
\]
(4)
\[
\begin{aligned}
(a+b+c)^7 &= ((a+b) + c)^7 \\
&= (a+b)^7 + 7 (a+b)^6 c + 21 (a+b)^5 c^2 + \cdots
\end{aligned}
\]
\[
\begin{aligned}
21 (a+b)^5 c^2 = 21(a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5)c^2
\end{aligned}
\]
Hence the coefficient of \(a^2b^3c^2\) is \(21 \times 10 = 210\).
(5)
\[
\begin{aligned}
(2a+b+c)^7 &= ((2a+b) + c)^7 \\
&= (2a+b)^7 + 7 (2a+b)^6 c + 21 (2a+b)^5 c^2 + \cdots
\end{aligned}
\]
\[
\begin{aligned}
21 (2a+b)^5 c^2 = 21(2^5a^5 + 5 \times 2^4a^4 b + 10 \times 2^3 a^3 b^2 + 10 \times 2^2 a^2 b^3 + 5 \times 2 a b^4 + b^5)c^2
\end{aligned}
\]
Hence the coefficient of \(a^2b^3c^2\) is \(21 \times 10 \times 2^2 = 840\).