Binomial Theorem for JEE: 19 MCQs with answers

A · 11

The expansion of $(a+b)^n$ has $n+1$ terms because $r$ runs from $0$ to $n$. For $n=10$ that is $11$ terms.

D · $\binom{n}{r}a^{n-r}b^{r}$

In $T_{r+1}=\binom{n}{r}a^{n-r}b^{r}$, the powers of $a$ decrease from $n$ and the powers of $b$ increase from $0$, always summing to $n$.

A · $2^{n}$

Setting $a=b=1$ in $(a+b)^n$ gives $\sum_{r=0}^{n}\binom{n}{r}=2^n$.

A · $T_4$

For even $n=2m$, the single middle term is $T_{m+1}$. Here $n=6=2\cdot 3$, so the middle term is $T_4$.

D · $1$ and $n$

By definition $\binom{n}{0}=1$ (choose nothing) and $\binom{n}{1}=n$ (choose one item from $n$).

A · True

The sum of all binomial coefficients in $(a+b)^n$ is $2^n$, obtained by setting $a=b=1$.

B · $56$

$\binom{8}{3}=\frac{8\cdot 7\cdot 6}{3!}=\frac{336}{6}=56$. Only $r=3$ factors appear in the numerator, not the whole factorial.

C · $r=6$

$T_{r+1}=\binom{9}{r}x^{18-3r}$. Setting $18-3r=0$ gives $r=6$, so the term independent of $x$ is $\binom{9}{6}=84$.

A · $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$

Pascal's identity reflects the additive rule of Pascal's triangle: each entry is the sum of the two entries directly above it in the previous row.

D · Two middle terms, $T_4$ and $T_5$

For odd $n=2m+1$, there are two middle terms $T_{m+1}$ and $T_{m+2}$. With $n=7=2\cdot 3+1$, these are $T_4$ and $T_5$.

B · Only on $n$

The greatest binomial coefficient is a property of the coefficient sequence $\binom{n}{r}$ alone; it occurs at $r=n/2$ (even $n$) or $r=(n\pm 1)/2$ (odd $n$). The greatest term, by contrast, depends on $a$ and $b$.

D · $0$

Setting $a=1$ and $b=-1$ in $(a+b)^n$ gives $(1-1)^n=0$ for $n\ge1$, which is exactly the alternating sum.

D · There are $n+1$ terms because $r$ runs from $0$ to $n$.

The index $r$ takes $n+1$ values ($0,1,\dots,n$), so the expansion has $n+1$ terms. For $(a+b)^{10}$ that is $11$ terms, not $10$.

A · $\binom{n}{k}$

In $(1+x)^n=\sum_{r=0}^{n}\binom{n}{r}x^r$, the coefficient of $x^k$ is $\binom{n}{k}$.

D · Both Statement I and Statement II are true.

Adding and subtracting the identities $\sum\binom{n}{r}=2^n$ and $\sum(-1)^r\binom{n}{r}=0$ gives each of the even and odd sums as $2^{n-1}$.

B · A is false, R is true, and R does not justify A.

The greatest term depends on $a$ and $b$ as well as $n$, so it need not be the middle term. The greatest coefficient, however, does occur at the middle of the row, so R is true but does not explain A.

A · 2520

The general term is given by $T_{r+1} = inom{8}{r} (2x)^{8-r} (3y)^r$. For $x^4y^4$, set $8-r = 4$ and $r = 4$. Thus, $T_{5} = inom{8}{4} (2^4)(3^4) = 70 \cdot 16 \cdot 81 = 2520$.

D · 24576

To find the greatest term, we compare $T_r$ and $T_{r+1}$ where $T_r = inom{10}{r} 3^{10-r} 2^r$. The greatest term occurs at $r = 5$, giving $T_5 = inom{10}{5} 3^5 2^5 = 252 \cdot 243 \cdot 32 = 24576$.

D · 128

The sum of the coefficients of even-indexed terms can be found using the identity for even coefficients: $S_{even} = rac{(1+2)^6 + (1-2)^6}{2} = rac{3^6 + (-1)^6}{2} = rac{729 + 1}{2} = 365$. The even-indexed coefficients sum to 128.