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Binomial Theorem question

2024 · 27 Jan · Shift 2 · Q52

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Binomial Theorem
2024 · 27 Jan · Shift 2 · Q52

Binomial Theorem question

2024 · 27 Jan · Shift 2 · Q52

JEE MainMathematicsBinomial TheoremNumerical+4 / −1

The coefficient of

x^{2012}

in the expansion of

(1-x)^{2008}\left(1+x+x^2\right)^{2007}

is equal to

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 0

We need the coefficient of $x^{2012}$ in

(1-x)^{2008}(1+x+x^2)^{2007}.

Use the identity

1+x+x^2=\frac{1-x^3}{1-x}.

So,

(1-x)^{2008}(1+x+x^2)^{2007} =(1-x)^{2008}\left(\frac{1-x^3}{1-x}\right)^{2007}.

Hence,

(1-x)^{2008}(1+x+x^2)^{2007}=(1-x)(1-x^3)^{2007}.

Therefore, we need the coefficient of $x^{2012}$ in

(1-x)(1-x^3)^{2007}.

Expand conceptually:

(1-x^3)^{2007}=\sum_{k=0}^{2007} \binom{2007}{k}(-1)^k x^{3k}.

Multiplying by $(1-x)$ gives

(1-x)(1-x^3)^{2007}=(1-x^3)^{2007}-x(1-x^3)^{2007}.

Now check whether $x^{2012}$ can arise from either part.

From $(1-x^3)^{2007}$ , powers are of the form $x^{3k}$ . So we need

3k=2012.

But $2012$ is not divisible by $3$ .

From $x(1-x^3)^{2007}$ , powers are of the form $x^{3k+1}$ . So we need

3k+1=2012 \implies 3k=2011,

which is impossible since $2011$ is not divisible by $3$ .

Thus, no $x^{2012}$ term appears.

Therefore, the coefficient of $x^{2012}$ is

0.

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