If,(1+x)^n=a0+a1x+a2x^2+........anx^n
Then show that,
1/3(2^n+2cosnπ\3)=a0+a3+a6+........
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Answer:
put =ωandnotethat1+ω+ω
2
=0
where ω=−
2
1
+i
2
3
andω
2
=−
2
1
−i
2
3
Δ=(a
0
+a
3
+a
6
+......)+ω(a
1
+a
4
+......)+ω(a
2
+a
3
......)
or Δ=Aω+Bω
2
+C=0
equating real and imaginary part of both sides
A−
2
B
−
2
C
=0
2
3
(B−C)=0
B = C and hence A = B
A \, = \, B \, = \., C
Again putting x = 1 in the given relation . we get 3^ n = sum of the all the condition = A + B + C
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