Math, asked by Anonymous, 10 months ago

solve the above attachment ⬆️⬆️⬆️


don't dare to spam❎❎❎​

Attachments:

Answers

Answered by vbhai97979
9

Answer:

3. Given, (1 + x)n = a0 + a1 x + a2 x2 +.......... + an xn .............1

Put x = 1, we get

(1 + 1)n = a0 + a1 + a2 +.......... + an

=> 2n = a0 + a1 + a2 +.......... + an ...........2

Again, put x = w and x = w2 in equation 2, we get

(1 + w)n = a0 + a1 *w + a2 * w2 +.......... + an * wn

=> (1 + w)n = (a0 + a3 ......) + w(a1 + a4 +......) + w2 (a2 + a5 + .......) + ........... .................3

and (1 + w2 )n = (a0 + a3 ......) + w2 (a1 + a4 +......) + w (a2 + a5 + .......) + ........... ................4

Add equation 2, 3 and 4, we get

2n + (1 + w)n + (1 + w2 )n = 3(a0 + a3 + a6 + .........)

=> 2n + (-w2 )n + (-w)n = 3(a0 + a3 + a6 + .........) {since 1 + w + w2 = 0 and w3 = 1 }

=> 2n + {-(-1/2 - √3/2 )}n + {-(-1/2 + √3/2 )}n = 3(a0 + a3 + a6 + .........) {since w = (-1/2 + √3/2 ) and w2 = (-1/2 - √3/2 )}

=> 2n + {-(-1/2 - √3/2 )}n + {1/2 - √3/2 )}n = 3(a0 + a3 + a6 + .........)

=> 2n + {cos π/3 + i * sin π/3}n + {cos π/3 - i * sin π/3}n = 3(a0 + a3 + a6 + .........)

=> 2n + cos nπ/3 + i * sin nπ/3 + cos nπ/3 - i * sin nπ/3 = 3(a0 + a3 + a6 + .........)

=> 2n + 2*cos nπ/3 = 3(a0 + a3 + a6 + .........)

=> a0 + a3 + a6 + ......... = (2n + 2*cos nπ/3)/3

Answered by Anonymous
2

\mathbb\pink{\underline{Answer}}

Given,

→ (1 + x)n = a0 + a1 x + a2 x2 +.......... + an xn .............1

Put x = 1, we get

→(1 + 1)n = a0 + a1 + a2 +.......... + an

→ 2n = a0 + a1 + a2 +.......... + an ...........2

Again, put x = w and x = w2 in equation 2, we get

→(1 + w)n = a0 + a1 *w + a2 * w2 +.......... + an * wn

→ (1 + w)n = (a0 + a3 ......) + w(a1 + a4 +......) + w2 (a2 + a5 + .......) + ........... .................3

and (1 + w2 )n = (a0 + a3 ......) + w2 (a1 + a4 +......) + w (a2 + a5 + .......) + ........... ................4

Add equation 2, 3 and 4, we get

→2n + (1 + w)n + (1 + w2 )n = 3(a0 + a3 + a6 + .........)

→ 2n + (-w2 )n + (-w)n = 3(a0 + a3 + a6 + .........) {since 1 + w + w2 = 0 and w3 = 1 }

→ 2n + {-(-1/2 - √3/2 )}n + {-(-1/2 + √3/2 )}n = 3(a0 + a3 + a6 + .........) {since w = (-1/2 + √3/2 ) and w2 = (-1/2 - √3/2 )}

→2n + {-(-1/2 - √3/2 )}n + {1/2 - √3/2 )}n = 3(a0 + a3 + a6 + .........)

→ 2n + {cos π/3 + i * sin π/3}n + {cos π/3 - i * sin π/3}n = 3(a0 + a3 + a6 + .........)

→ 2n + cos nπ/3 + i * sin nπ/3 + cos nπ/3 - i * sin nπ/3 = 3(a0 + a3 + a6 + .........)

→ 2n + 2*cos nπ/3 = 3(a0 + a3 + a6 + .........)

→a0 + a3 + a6 + ......... = (2n + 2*cos nπ/3)/3

Similar questions