展開する (2sqrt(3)+j^2)^4

解

展開する

(23 + j^{2})^{4}

144 + 963 j^{2} + 72 j^{4} + 83 j^{6} + j^{8}

解答ステップ

(23 + j^{2})^{4}

2項定理を適用する:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 23, b = j^{2}

= i = 0 \sum 4 (i 4) (23)^{(4 - i)} (j^{2})^{i}

総和を展開する

= \frac{4 !}{0 ! ( 4 - 0 ) !} (23)^{4} (j^{2})^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (23)^{3} (j^{2})^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (23)^{2} (j^{2})^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (23)^{1} (j^{2})^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (23)^{0} (j^{2})^{4}

簡素化

\frac{4 !}{0 ! ( 4 - 0 ) !} (23)^{4} (j^{2})^{0} : 144

簡素化

\frac{4 !}{1 ! ( 4 - 1 ) !} (23)^{3} (j^{2})^{1} : 963 j^{2}

簡素化

\frac{4 !}{2 ! ( 4 - 2 ) !} (23)^{2} (j^{2})^{2} : 72 j^{4}

簡素化

\frac{4 !}{3 ! ( 4 - 3 ) !} (23)^{1} (j^{2})^{3} : 83 j^{6}

簡素化

\frac{4 !}{4 ! ( 4 - 4 ) !} (23)^{0} (j^{2})^{4} : j^{8}

= 144 + 963 j^{2} + 72 j^{4} + 83 j^{6} + j^{8}