Math Question and Answers


Question and Answers Forum

All Questions Topic List
Mensuration Questions
Previous in All Question Next in All Question
Previous in Mensuration Next in Mensuration
Question Number 76663 by aliesam last updated on 29/Dec/19

	Answered by benjo 1/2 santuyy last updated on 29/Dec/19

	$2^{3 n} ?$
	Answered by mind is power last updated on 29/Dec/19

	${(1 + j)}^{3 n} = Σ C_{3 n}^{k} j^{k} = \underset{k = 0}{\sum^{n}} C_{3 n}^{3 k} + j \underset{k = 0}{\sum^{n - 1}} C_{3 n}^{3 k + 1} + j^{2} \underset{k = 0}{\sum^{n - 1}} C_{3 n}^{3 k + 2} . . . A$ $withe j = e^{\frac{2 i π}{3}}$ ${(1 + j^{2})}^{3 n} = Σ C_{3 n}^{k} j^{2 k} = j^{2} \underset{k = 0}{\sum^{n - 1}} C_{3 n}^{3 k + 1} + j \underset{k = 0}{\sum^{n - 1}} C_{3 n}^{3 k + 2} + \underset{k = 0}{\sum^{n}} C_{3 n}^{3 k} . . . B$ ${(1 + 1)}^{3 n} = \underset{k = 0}{\sum^{3 n}} C_{3 n}^{k} = \underset{k = 0}{\sum^{n}} C_{3 n}^{3 k} + \underset{k = 0}{\sum^{n - 1}} C_{3 n}^{3 k + 1} + \underset{k = 0}{\sum^{n - 1}} C_{3 n}^{3 k + 2}$ $1 + j + j^{2} = 0 \Rightarrow$ $A + B \Leftrightarrow {(1 + j)}^{3 n} + {(1 + j^{2})}^{3 n} = 2 \underset{k = 0}{\sum^{n}} C_{3 n}^{3 k} - Σ C_{3 n}^{3 k + 1} - Σ C_{3 n}^{3 k + 2}$ $A - B \Rightarrow {(1 + j)}^{3 n} - {(1 + j^{2})}^{3 n} = (j - j^{2}) Σ C_{3 n}^{3 k + 1} - (j - j^{2}) Σ C_{3 n}^{3 k + 2}$ $X = Σ C_{3 n}^{3 k}, Y = Σ C_{3 n}^{3 k + 1}, Z = Σ C_{3 n}^{3 k + 2}$ $x + y + z = 2^{3 n}$ $2 x - y - z = {(1 + j)}^{3 n} + {(1 + j^{2})}^{3 n}$ $3 x = \frac{2^{3 n} + {(1 + j)}^{3 n} + {(1 + j^{2})}^{3 n}}{}$ ${(1 + j)}^{3 n} = {(- j^{2})}^{3 n} = {(- 1)}^{n}$ ${(1 + j^{2})}^{3 n} = {(- j)}^{3 n} = {(- 1)}^{n}$ $\Rightarrow x = \frac{2^{3 n} + 2 {(- 1)}^{n}}{3} = \underset{k = 0}{\sum^{n}} C_{3 n}^{3 k}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum