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Question
If f(x) = `|((1 + x)^17, (1 + x)^19, (1 + x)^23),((1 + x)^23, (1 + x)^29, (1 + x)^34),((1 +x)^41, (1 +x)^43, (1 + x)^47)|` = A + Bx + Cx2 + ..., then A = ______.
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Solution
If f(x) = `|((1 + x)^17, (1 + x)^19, (1 + x)^23),((1 + x)^23, (1 + x)^29, (1 + x)^34),((1 +x)^41, (1 +x)^43, (1 + x)^47)|` = A + Bx + Cx2 + ..., then A = 0.
Explanation:
Given that `|((1 + x)^17, (1 + x)^19, (1 + x)^23),((1 + x)^23, (1 + x)^29, (1 + x)^34),((1 +x)^41, (1 +x)^43, (1 + x)^47)|` = A + Bx + Cx2 + ...
Taking (1 + x)17, (1 + x)23 and (1 + x)41 common from R1, R2 and R3 respectively
`(1 + x)^17 * (1 + x)^23 * (1 + x)^41 |(1, (1 + x)^2, (1 x)^6),(1, (1 + x)^6, (1 + x)^11),(1, (1 + x)^2, (1 + x)^6)|`
`(1 + x)^17 * (1 + x)^23 * (1 + x)^41 * 0` ....(R1 and R3 are identical)
∴ 0 = A + Bx + Cx2 + …
By comparing the like terms, we get A = 0.
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