If the Coefficient of x^30 in the expansion of (1+\frac1x)^6(1+x^2)^7(1-x^3)^8 ; x eq 0 is \alpha , then |\alpha| equals \underline\hspace2cm .

Binomial Theorem PYQ — JEE Main 2024 · 1 Feb · Shift 1 · Q53 with Solution

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Correct answer: 678

We need the coefficient of $x^{30}$ in

\left(1+\frac{1}{x}\right)^6(1+x^2)^7(1-x^3)^8.

Expand each factor using the binomial theorem:

\left(1+\frac{1}{x}\right)^6=\sum_{i=0}^6 \binom{6}{i}x^{-i},

(1+x^2)^7=\sum_{j=0}^7 \binom{7}{j}x^{2j},

(1-x^3)^8=\sum_{k=0}^8 \binom{8}{k}(-1)^k x^{3k}.

So the general term in the product is

\binom{6}{i}\binom{7}{j}\binom{8}{k}(-1)^k x^{-i+2j+3k}.

We want the power of $x$ to be $30$ , so

-i+2j+3k=30,

with

0\le i\le 6,\quad 0\le j\le 7,\quad 0\le k\le 8.

Now find integer solutions of

-i+2j+3k=30.

Since $2j\le 14$ and $-i\le 0$ , we need large $k$ .

Try $k=8$ :

-i+2j+24=30 \implies 2j-i=6.

Now check $j=0$ to $7$ with $i=2j-6$ in $[0,6]$ :

$j=3 \implies i=0$
$j=4 \implies i=2$
$j=5 \implies i=4$
$j=6 \implies i=6$
$j=7 \implies i=8$ not allowed

So valid solutions for $k=8$ are:

(i,j,k)=(0,3,8),(2,4,8),(4,5,8),(6,6,8).

Try $k=7$ :

-i+2j+21=30 \implies 2j-i=9.

Since $2j\le 14$ and $i\ge 0$ , possibilities are:

$j=5 \implies i=1$
$j=6 \implies i=3$
$j=7 \implies i=5$

So valid solutions are:

(i,j,k)=(1,5,7),(3,6,7),(5,7,7).

Try $k\le 6$ :

-i+2j+3k\le 0+14+18=32,

but checking $k=6$ gives

2j-i=12,

possible only if $j=6,i=0$ or $j=7,i=2$ ? Wait:

$j=6 \implies i=0$ gives valid
$j=7 \implies i=2$ gives valid So we must include $k=6$ too.

For $k=6$ :

-i+2j+18=30 \implies 2j-i=12.

Valid solutions:

$j=6 \implies i=0$
$j=7 \implies i=2$

So valid solutions are:

(i,j,k)=(0,6,6),(2,7,6).

For $k=5$ :

-i+2j+15=30 \implies 2j-i=15.

But $2j\le 14$ , impossible. So no more solutions.

Now compute the coefficient contribution from each valid triple. The coefficient is

\alpha=\sum \binom{6}{i}\binom{7}{j}\binom{8}{k}(-1)^k.

Since $\binom{8}{8}=1$ , $\binom{8}{7}=8$ , $\binom{8}{6}=28$ :

For $k=8$ (positive sign):

\binom{6}{0}\binom{7}{3}+\binom{6}{2}\binom{7}{4}+\binom{6}{4}\binom{7}{5}+\binom{6}{6}\binom{7}{6}

=1\cdot 35+15\cdot 35+15\cdot 21+1\cdot 7

=35+525+315+7=882.

For $k=7$ (negative sign):

-8\left[\binom{6}{1}\binom{7}{5}+\binom{6}{3}\binom{7}{6}+\binom{6}{5}\binom{7}{7}\right]

=-8\left[6\cdot 21+20\cdot 7+6\cdot 1\right]

=-8(126+140+6)=-8\cdot 272=-2176.

For $k=6$ (positive sign):

28\left[\binom{6}{0}\binom{7}{6}+\binom{6}{2}\binom{7}{7}\right]

=28(1\cdot 7+15\cdot 1)=28(22)=616.

Therefore,

\alpha=882-2176+616=-678.

Hence,

|\alpha|=678.

Comparison with stored answer: Stored correct answer = $678$ . Our derived answer is also $678$ , so they agree.

More from Binomial Theorem

View written solutionFree

Correct answer: 678

We need the coefficient of $x^{30}$ in

\left(1+\frac{1}{x}\right)^6(1+x^2)^7(1-x^3)^8.

Expand each factor using the binomial theorem:

\left(1+\frac{1}{x}\right)^6=\sum_{i=0}^6 \binom{6}{i}x^{-i},

(1+x^2)^7=\sum_{j=0}^7 \binom{7}{j}x^{2j},

(1-x^3)^8=\sum_{k=0}^8 \binom{8}{k}(-1)^k x^{3k}.

So the general term in the product is

\binom{6}{i}\binom{7}{j}\binom{8}{k}(-1)^k x^{-i+2j+3k}.

We want the power of $x$ to be $30$ , so

-i+2j+3k=30,

with

0\le i\le 6,\quad 0\le j\le 7,\quad 0\le k\le 8.

Now find integer solutions of

-i+2j+3k=30.

Since $2j\le 14$ and $-i\le 0$ , we need large $k$ .

Try $k=8$ :

-i+2j+24=30 \implies 2j-i=6.

Now check $j=0$ to $7$ with $i=2j-6$ in $[0,6]$ :

$j=3 \implies i=0$
$j=4 \implies i=2$
$j=5 \implies i=4$
$j=6 \implies i=6$
$j=7 \implies i=8$ not allowed

So valid solutions for $k=8$ are:

(i,j,k)=(0,3,8),(2,4,8),(4,5,8),(6,6,8).

Try $k=7$ :

-i+2j+21=30 \implies 2j-i=9.

Since $2j\le 14$ and $i\ge 0$ , possibilities are:

$j=5 \implies i=1$
$j=6 \implies i=3$
$j=7 \implies i=5$

So valid solutions are:

(i,j,k)=(1,5,7),(3,6,7),(5,7,7).

Try $k\le 6$ :

-i+2j+3k\le 0+14+18=32,

but checking $k=6$ gives

2j-i=12,

possible only if $j=6,i=0$ or $j=7,i=2$ ? Wait:

$j=6 \implies i=0$ gives valid
$j=7 \implies i=2$ gives valid So we must include $k=6$ too.

For $k=6$ :

-i+2j+18=30 \implies 2j-i=12.

Valid solutions:

$j=6 \implies i=0$
$j=7 \implies i=2$

So valid solutions are:

(i,j,k)=(0,6,6),(2,7,6).

For $k=5$ :

-i+2j+15=30 \implies 2j-i=15.

But $2j\le 14$ , impossible. So no more solutions.

Now compute the coefficient contribution from each valid triple. The coefficient is

\alpha=\sum \binom{6}{i}\binom{7}{j}\binom{8}{k}(-1)^k.

Since $\binom{8}{8}=1$ , $\binom{8}{7}=8$ , $\binom{8}{6}=28$ :

For $k=8$ (positive sign):

\binom{6}{0}\binom{7}{3}+\binom{6}{2}\binom{7}{4}+\binom{6}{4}\binom{7}{5}+\binom{6}{6}\binom{7}{6}

=1\cdot 35+15\cdot 35+15\cdot 21+1\cdot 7

=35+525+315+7=882.

For $k=7$ (negative sign):

-8\left[\binom{6}{1}\binom{7}{5}+\binom{6}{3}\binom{7}{6}+\binom{6}{5}\binom{7}{7}\right]

=-8\left[6\cdot 21+20\cdot 7+6\cdot 1\right]

=-8(126+140+6)=-8\cdot 272=-2176.

For $k=6$ (positive sign):

28\left[\binom{6}{0}\binom{7}{6}+\binom{6}{2}\binom{7}{7}\right]

=28(1\cdot 7+15\cdot 1)=28(22)=616.

Therefore,

\alpha=882-2176+616=-678.

Hence,

|\alpha|=678.

Comparison with stored answer: Stored correct answer = $678$ . Our derived answer is also $678$ , so they agree.

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