Question:
(1+ x)^n=a0 +a1x+ a2x^2 ..+.......... +anx^n then find i)ao-a2 +a4-a6 .......... ii)a1-a3+ a5-a7 ......... iii)a0+ a3+ a6 +a9+-------
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
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