First slide

Binomial theorem for positive integral Index

Question

If the coefficient of xⁿ in $(1 + x)^{101} {(1 - x + x^{2})}^{100}$ is nonzero, then n cannot be of the form

Moderate

A

3r + 1
B

3r
C

3r + 2
D

3r-2

Solution

We have,
$\begin{array}{l} (1 + x)^{101} {(1 - x + x^{2})}^{100} \\ = (1 + x) {((1 + x) (1 - x + x^{2}))}^{100} \\ = (1 + x) {(1 + x^{3})}^{100} \\ = (1 + x) \{C_{0} + C_{1} x^{3} + C_{2} x^{6} + \dots + C_{100} x^{300}\}, where C_{r} =^{100} C_{r} \\ = (1 + x) \sum_{r = 0}^{n}^{n} C_{r} x^{3 r} = \sum_{r = 0}^{n}^{n} C_{r} x^{3 r} + \sum_{r = 0}^{n}^{n} C_{r} x^{3 r + 1} \end{array}$
Hence, there will be no term containing 3r + 2.

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