The constant term in the expansion of ≤ft(1+\;1/x)²0} (30 x(1+x)²9}+(1+x)³0}) is

Correct Answer is : 50C20+30⋅49C19{ }^{50} C ₂0}+30 · { }^{49} C ₁9}50C20+30⋅49C19

{ }^{50} C ₂0}+30 · { }^{50} C ₂9}

{ }^{50} C ₁9}+30 · { }^{49} C ₁9}

{ }^{50} C ₂0}+30 · { }^{49} C ₂0}

{ }^{50} C ₂0}+30 · { }^{49} C ₁9}

Correct Answer is : 50C20+30⋅49C19{ }^{50} C ₂0}+30 · { }^{49} C ₁9}50C20+30⋅49C19

Test Index

TG EAMCET 2 May 2025 Shift 2 Paper

Question : 19 of 160

Marks: +1, -0

\left(1+\;\frac{1}{x}\right)^{20} (30 x(1+x)^{29}+(1+x)^{30})

Solution:

Let the given expression,

E= \left(1+\;\frac{1}{x}\right)^{20} [30 x(1+x)^{29}+(1+x)^{30}]

Here,

30 x(1+x)^{29}+(1+x)^{30}

\;=(1+x)^{29}[30 x+(1+x)]

\;=(1+x)^{29}(31 x+1)

\therefore \;\; E\;= \left(1+\;\frac{1}{x}\right)^{20}(1+x)^{29}(31 x+1)

\;=\;\frac{(x+1)^{20}}{x^{20}}(1+x)^{29}(31 x+1)

\;=\;\frac{(x+1)^{49}}{x^{20}}(31 x+1)

\;=(x+1)^{49} (31 x^{-19}+x^{-20})

\;=31 x^{-19}(x+1)^{49}+x^{-20}(x+1)^{49}

For constant term

x^0

\therefore

Constant term in

31 x^{-19}(x+1)^{49}

i.e.,

x^{-19} \cdot x^k=x^0 \Rightarrow k=19

\therefore

Constant term

=31{ }^{49} C_{19}

Constant term in

x^{-20}(x+1)^{49}

i.e.,

x^{-20} \cdot x^k=x^0 \Rightarrow k=20

\therefore

Constant term

=1 \times { }^{49} C_{20}

\therefore

Sum of constant term

=31{ }^{49} C_{19}+{ }^{49} C_{20}

=30^{49} C_{19}+{ }^{49} C_{19}+{ }^{49} C_{20}

=30 \cdot { }^{49} C_{19}+{ }^{50} C_{20}

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