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Method 1:
(a+2x)^5
=(5C0)(a^5)(2x)^0 + (5C1)(a^4)(2x)^1 + (5C2)(a^3)(2x)^2 + (5C3)(a^2)(2x)^3 + (5C4)(a^1)(2x)^4 + (5C5)(a^0)(2x)^5
=a^5 + (10a^4)x + (40a^3)x^2 +(80a^2)x^3 + 80ax^4 + 32x^5
so,
40a^3=40
a^3=1
a=1
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Method 2:
the (r+1)th term
=(5Cr)(a^r)[(2x)^(5-r)]
=(5Cr)(a^r)[2^(5-r)][x^(5-r)]
To find the coefficient of x^2
5-r=2 , i.e. r=3
So the coefficient of x^2
=(5C3)(a^3)[2^(5-3)]
=(10)(a^3)(4)
=40a^3
so,
40a^3=40
a^3=1
a=1 |
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