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in Binomial theorem by (52.7k points)

The value of loge (1 + ax2 + a2 + a/x2) is

(A)  a(x2  - 1/x2) - a2/2(x4 - 1/x4) + a3/3(x6 - 1/x6)  -....

(B)  a(x2  + 1/x2) - a2/2(x4 + 1/x4) + a3/3(x6 + 1/x6)  -....

(C)  a(x2  + 1/x2) + a2/2(x4 + 1/x4) + a3/3(x6 + 1/x6) + ....

(D)  a(x2  - 1/x2) + a2/2(x4 - 1/x4) + a3/3(x6 - 1/x6) +....

1 Answer

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Best answer

Correct option (B)  a(x2  + 1/x2) - a2/2(x4 + 1/x4) + a3/3(x6 + 1/x6)  -....

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