The coefficient of x^70 in x^2(1+x)^98+x^3(1+x)^97+x^4(1+x)^96+\ldots+x^54(1+x)^46 is ^99 C_p- ^46 C_q . Then a possible value of p+q is :

Binomial Theorem PYQ — JEE Main 2024 · 9 Apr · Shift 1 · Q38 with Solution

View written solutionFree

Correct answer: B

Write the general term

The given sum is

S=x^2(1+x)^{98}+x^3(1+x)^{97}+x^4(1+x)^{96}+\cdots +x^{54}(1+x)^{46}.

This can be written as

S=\sum_{r=2}^{54} x^r(1+x)^{100-r}.

Indeed, for $r=2$ , exponent of $(1+x)$ is $98$ , and for $r=54$ , it is $46$ .

Find the coefficient of $x^{70}$ in a general term

x^r(1+x)^{100-r},

to get $x^{70}$ , we need the coefficient of

x^{70-r}

in $(1+x)^{100-r}$ .

So the contribution from the $r$ th term is

\binom{100-r}{70-r}.

Hence total coefficient is

\sum_{r=2}^{54} \binom{100-r}{70-r}.

Using symmetry of binomial coefficients,

\binom{100-r}{70-r}=\binom{100-r}{30}.

Therefore the required coefficient is

\sum_{r=2}^{54} \binom{100-r}{30}.

Let $k=100-r$ . Then as $r$ goes from $2$ to $54$ , $k$ goes from $98$ down to $46$ . So

\sum_{r=2}^{54} \binom{100-r}{30}= \sum_{k=46}^{98} \binom{k}{30}.

Use the hockey-stick identity

We use

\sum_{k=m}^{n} \binom{k}{m}=\binom{n+1}{m+1}.

Thus,

\sum_{k=46}^{98} \binom{k}{30} =\sum_{k=30}^{98} \binom{k}{30}-\sum_{k=30}^{45} \binom{k}{30}

=\binom{99}{31}-\binom{46}{31}.

Now use symmetry again:

\binom{99}{31}=\binom{99}{68}, \qquad \binom{46}{31}=\binom{46}{15}.

So the coefficient is of the form

{}^{99}C_p-{}^{46}C_q

with one possible choice

p=68,\qquad q=15.

Hence

p+q=68+15=83.

Check the options

The possible value is

83.

So the correct option is B.

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.

More from Binomial Theorem

View written solutionFree

Correct answer: B

Write the general term

The given sum is

S=x^2(1+x)^{98}+x^3(1+x)^{97}+x^4(1+x)^{96}+\cdots +x^{54}(1+x)^{46}.

This can be written as

S=\sum_{r=2}^{54} x^r(1+x)^{100-r}.

Indeed, for $r=2$ , exponent of $(1+x)$ is $98$ , and for $r=54$ , it is $46$ .

Find the coefficient of $x^{70}$ in a general term

x^r(1+x)^{100-r},

to get $x^{70}$ , we need the coefficient of

x^{70-r}

in $(1+x)^{100-r}$ .

So the contribution from the $r$ th term is

\binom{100-r}{70-r}.

Hence total coefficient is

\sum_{r=2}^{54} \binom{100-r}{70-r}.

Using symmetry of binomial coefficients,

\binom{100-r}{70-r}=\binom{100-r}{30}.

Therefore the required coefficient is

\sum_{r=2}^{54} \binom{100-r}{30}.

Let $k=100-r$ . Then as $r$ goes from $2$ to $54$ , $k$ goes from $98$ down to $46$ . So

\sum_{r=2}^{54} \binom{100-r}{30}= \sum_{k=46}^{98} \binom{k}{30}.

Use the hockey-stick identity

We use

\sum_{k=m}^{n} \binom{k}{m}=\binom{n+1}{m+1}.

Thus,

\sum_{k=46}^{98} \binom{k}{30} =\sum_{k=30}^{98} \binom{k}{30}-\sum_{k=30}^{45} \binom{k}{30}

=\binom{99}{31}-\binom{46}{31}.

Now use symmetry again:

\binom{99}{31}=\binom{99}{68}, \qquad \binom{46}{31}=\binom{46}{15}.

So the coefficient is of the form

{}^{99}C_p-{}^{46}C_q

with one possible choice

p=68,\qquad q=15.

Hence

p+q=68+15=83.

Check the options

The possible value is

83.

So the correct option is B.

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.

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