The-number-of-terms-in-the-expansion-of-1-2x-x-2-20-when-expanded-in-descending-powers-of-x-is-

Question Number 10544 by j.masanja06@gmail.com last updated on 17/Feb/17

The number of terms in the expansion of

{(1 + 2 x + x^{2})}^{20} when expanded in descending

powers of x, is

Commented by FilupS last updated on 18/Feb/17

x^{2} + 2 x + 1 = {(x + 1)}^{2}

\Rightarrow {(1 + 2 x + x^{2})}^{20} = {(x + 1)}^{40}

{(a + b)}^{n} = \underset{u = 0}{\sum^{n}} (\begin{matrix} n \\ u \end{matrix}) a^{n - u} b^{n}

a = 1, b = x

{(x + 1)}^{40} = \underset{u = 0}{\sum^{40}} (\begin{matrix} 40 \\ u \end{matrix}) x^{40 - u}

(\begin{matrix} 40 \\ u \end{matrix}) = \frac{40!}{u! \cdot (40 - u)!}

{(x + 1)}^{40} = 1 + \underset{u = 0}{\sum^{39}} (\begin{matrix} 40 \\ u \end{matrix}) x^{40 - u}

∴ there are 41 terms

\forall u \in (0, 39) \Rightarrow x^{40 - u} > 1

∴ there are 40 coefficients of x \to x^{40}

i . e . {(x + 1)}^{40} = a_{0} x^{40} + a_{1} x^{39} + \dots + a_{38} x^{2} + a_{39} x + 1

continue

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