The numerical value of the third term in the expression (x + y)^8, if x = 0.3 and y = 0.7 is 0.010
The correct option is a. 0.010
To determine the numerical value of the third term, we will first expand the given expression (x + y)^8 using the binomial theorem
From the binomial theorem, we have that,
\((x+y)^{n} =\left(\begin{array}{ccc}n\\0\\\end{array}\right)x^{n} y^{0} + \left(\begin{array}{ccc}n\\1\\\end{array}\right)x^{n-1} y^{1} + \left(\begin{array}{ccc}n\\2\\\end{array}\right)x^{n-2} y^{2} + \left(\begin{array}{ccc}n\
\\\end{array}\right)x^{0} y^{n}\)
where \(\left(\begin{array}{ccc}n\\0\\\end{array}\right) = ^{n} C_{r} = \frac{n!}{r!(n-r)!}\)
Then the expansion of the expression (x + y)^8 will give
\((x+y)^{8} = x^{8}+8x^{7}y+ 28x^{6}y^{2}+56x^{5}y^{3}+70x^{4}y^{4}+56x^{3}y^{5}+28x^{2}y^{6}+8xy^{7} +y^{8}\)
From the expansion above, the third term is \(28x^{6}y^{2}\)
Now, for the numerical value of the third term if x = 0.3 and y = 0.7, we will put these values into the expression for the third term
\(28x^{6}y^{2} = 28\times (0.3)^{6}(0.7)^{2}\)
= 28 × 0.000729 × 0.49
= 0.01000188
≅ 0.010
∴ The numerical value ≅ 0.010'
Hence, the numerical value of the third term in the expression (x + y)^8, if x = 0.3 and y = 0.7 is 0.010
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