NCERT Exemplar Solutions Class 11 Maths Binomial Theorem — Fill in the Blanks and True or False Q25–40

NCERT Exemplar Problems, Class 11 Mathematics — Binomial Theorem, Fill in the Blanks and True or False questions 25 to 40. All 16 questions solved.

Nine fill-in-the-blank questions (25 to 33) and seven true-or-false statements (34 to 40). The blanks ask for a largest coefficient, a count of terms, a constant term, a position, a remainder, and a middle term. The true-or-false statements test the standard facts: the number of terms in $(a+b)^n$, the sum of the binomial coefficients, divisibility by writing a number as $(1\pm 8)^k$, and when a term free of $x$ exists.

One printed answer is wrong, Question 34, where the key says False but the statement is True. It is explained where it occurs.

Key insight. To find a remainder or last digits (33, 35, 38), write the number as a sum or difference near a multiple of the divisor, such as $25 = 26 – 1$ or $3^{400} = (10-1)^{200}$, and expand with the binomial theorem. Every term but the first one or two is then a multiple of the divisor and can be dropped.

Fill in the blanks in Questions 25 to 33.

Question 25

The largest coefficient in the expansion of $(1+x)^{30}$ is ____.

Solution. The coefficients ${}^{30}C_r$ rise to a maximum at the middle term and then fall. For $n = 30$ the middle is $r = 15$, so the largest coefficient is ${}^{30}C_{15} = 155117520$.

Question 26

The number of terms in the expansion of $(x+y+z)^n$ is ____.

Solution. Write $(x+y+z)^n = [x + (y+z)]^n = \sum_{r=0}^{n}{}^nC_r\,x^{n-r}(y+z)^r$. The term with $(y+z)^r$ expands into $r + 1$ terms. So the total is

$$\sum_{r=0}^{n}(r+1) = \frac{(n+1)(n+2)}{2}.$$

Check with $n = 2$: $x^2, y^2, z^2, xy, yz, zx$ is $6$ terms, and $\frac{3\cdot4}{2} = 6$.

Question 27

In the expansion of $\left(x^2 – \dfrac{1}{x^2}\right)^{16}$, the value of the constant term is ____.

This uses the general term, which is no longer part of the current NCERT Class 11 syllabus, but it stays in the Exemplar, so it is solved here.

Solution. $T_{r+1} = {}^{16}C_r\,x^{2(16-r)}\left(-x^{-2}\right)^r = {}^{16}C_r(-1)^r x^{32-4r}$. The power is zero when $r = 8$, and $(-1)^8 = 1$, so the constant term is ${}^{16}C_8 = 12870$.

Question 28

If the seventh terms from the beginning and the end in the expansion of $\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^n$ are equal, then $n$ equals ____.

Solution. The seventh term from the beginning is ${}^nC_6(2^{1/3})^{n-6}(3^{-1/3})^6$. The seventh term from the end is ${}^nC_6(2^{1/3})^6(3^{-1/3})^{n-6}$, found by reversing the order of the two parts. The binomial coefficients are equal, so setting the terms equal and dividing,

$$(2^{1/3})^{n-12} = (3^{-1/3})^{n-12}.$$

Two different positive bases ($2^{1/3} \neq 3^{-1/3}$) raised to the same power are equal only if that power is zero, so $n – 12 = 0$.

Question 29

The coefficient of $a^{-6}b^4$ in the expansion of $\left(\dfrac1a – \dfrac{2b}{3}\right)^{10}$ is ____.

Solution. $T_{r+1} = {}^{10}C_r\,a^{-(10-r)}\left(-\dfrac{2b}{3}\right)^r$. For $b^4$ we need $r = 4$, and then the power of $a$ is $-(10-4) = -6$, as required. The coefficient is

$${}^{10}C_4\left(-\frac23\right)^4 = 210\cdot\frac{16}{81} = \frac{3360}{81} = \frac{1120}{27}.$$

Question 30

The middle term in the expansion of $(a^3 + ba)^{28}$ is ____.

The middle term is no longer part of the current NCERT Class 11 syllabus, but it stays in the Exemplar, so it is solved here.

Solution. The expression is $(a^3 + ba)^{28}$ with second part $ba$, that is $b$ times $a$. With $n = 28$, the middle term is $T_{15}$:

$$T_{15} = {}^{28}C_{14}(a^3)^{14}(ba)^{14} = {}^{28}C_{14}\,a^{42}\cdot a^{14}b^{14} = {}^{28}C_{14}\,a^{56}b^{14}.$$

Question 31

The ratio of the coefficients of $x^p$ and $x^q$ in the expansion of $(1+x)^{p+q}$ is ____.

Solution. The coefficients are ${}^{p+q}C_p$ and ${}^{p+q}C_q$. Since ${}^{p+q}C_q = {}^{p+q}C_{(p+q)-q} = {}^{p+q}C_p$, they are equal and the ratio is $1:1$.

Question 32

The position of the term independent of $x$ in the expansion of $\left(\dfrac{\sqrt x}{3} + \dfrac{3}{2x^2}\right)^{10}$ is ____.

Solution. $T_{r+1} = {}^{10}C_r\left(\dfrac{\sqrt x}{3}\right)^{10-r}\left(\dfrac{3}{2x^2}\right)^r$, so the power of $x$ is $\dfrac{10-r}{2} – 2r$. Setting it to zero gives $10 – r = 4r$, so $r = 2$. The term is $T_3$, the third term. Its value is ${}^{10}C_2\cdot\dfrac{1}{3^8}\cdot\dfrac{9}{4} = \dfrac{45\cdot 9}{4\cdot 6561} = \dfrac{5}{324}$, though the question asks only for the position.

Question 33

If $25^{15}$ is divided by $13$, the remainder is ____.

Solution. Write $25 = 26 – 1$, where $26 = 2\times13$:

$$25^{15} = (26 – 1)^{15} = 26^{15} – {}^{15}C_1 26^{14} + \cdots – 1.$$

Every term except the last is a multiple of $26$, hence of $13$. So $25^{15} = 13m – 1 = 13(m-1) + 12$, and the remainder is $12$, not $-1$.

State which of the statements in Questions 34 to 40 is True or False.

Question 34

The sum of the series $\displaystyle\sum_{r=0}^{10}{}^{20}C_r$ is $2^{19} + \dfrac{{}^{20}C_{10}}{2}$.

Solution. Use the symmetry ${}^{20}C_r = {}^{20}C_{20-r}$. The full sum is $\sum_{r=0}^{20}{}^{20}C_r = 2^{20}$. The terms $r = 0$ to $9$ pair with $r = 20$ down to $11$, so

$$2^{20} = 2\sum_{r=0}^{9}{}^{20}C_r + {}^{20}C_{10} \;\Rightarrow\; \sum_{r=0}^{9}{}^{20}C_r = \frac{2^{20} – {}^{20}C_{10}}{2}.$$

Adding the $r = 10$ term back,

$$\sum_{r=0}^{10}{}^{20}C_r = \frac{2^{20} – {}^{20}C_{10}}{2} + {}^{20}C_{10} = 2^{19} + \frac{{}^{20}C_{10}}{2}.$$

Numerically both sides are $616666$. So the statement is True.

The printed key says False for this question, which is wrong. The direct check settles it: $\sum_{r=0}^{10}{}^{20}C_r = 616666$ and $2^{19} + \frac{184756}{2} = 524288 + 92378 = 616666$.

Question 35

The expression $7^9 + 9^7$ is divisible by $64$.

Solution. Use the hint: $7^9 = (8-1)^9 = -(1-8)^9$ and $9^7 = (1+8)^7$, so $7^9 + 9^7 = (1+8)^7 – (1-8)^9$. Expanding both to the second power of $8$:

$$(1+8)^7 = 1 + 7\cdot8 + 64k_1 = 57 + 64k_1, \qquad (1-8)^9 = 1 – 9\cdot8 + 64k_2 = -71 + 64k_2.$$

Hence $7^9 + 9^7 = 57 + 71 + 64(k_1 – k_2) = 128 + 64(k_1-k_2)$, which is a multiple of $64$. Direct check: $7^9 + 9^7 = 45136576 = 64\times705259$. True.

Question 36

The number of terms in the expansion of $[(2x + y^3)^4]^7$ is $8$.

Solution. $[(2x+y^3)^4]^7 = (2x+y^3)^{28}$, a binomial raised to the power $28$. It has $28 + 1 = 29$ terms, not $8$. False.

Question 37

The sum of coefficients of the two middle terms in the expansion of $(1+x)^{2n-1}$ is equal to ${}^{2n-1}C_n$.

Solution. The power $2n – 1$ is odd, so there are two middle terms, $T_n$ and $T_{n+1}$, with coefficients ${}^{2n-1}C_{n-1}$ and ${}^{2n-1}C_n$. By Pascal’s rule their sum is

$${}^{2n-1}C_{n-1} + {}^{2n-1}C_n = {}^{2n}C_n,$$

not ${}^{2n-1}C_n$. (Test $n = 2$: $(1+x)^3$ has middle coefficients $3 + 3 = 6 = {}^4C_2$, while ${}^3C_2 = 3$.) False.

Question 38

The last two digits of the number $3^{400}$ are $01$.

Solution. Write $3^{400} = (3^2)^{200} = 9^{200} = (10 – 1)^{200}$. Then

$$(10-1)^{200} = 1 – 200\cdot10 + {}^{200}C_2\cdot10^2 – \cdots$$

Every term after the first is a multiple of $100$ (the second is $2000$, the rest carry $10^2$ or higher). So $3^{400} \equiv 1 \pmod{100}$, and the last two digits are $01$. True.

Question 39

If the expansion of $\left(x – \dfrac{1}{x^2}\right)^{2n}$ contains a term independent of $x$, then $n$ is a multiple of $2$.

This uses the general term, which is no longer part of the current NCERT Class 11 syllabus, but it stays in the Exemplar, so it is solved here.

Solution. $T_{r+1} = {}^{2n}C_r\,x^{2n-r}(-1)^r x^{-2r} = (-1)^r\,{}^{2n}C_r\,x^{2n-3r}$. A term independent of $x$ needs $2n = 3r$, so $r = \dfrac{2n}{3}$ must be a whole number. Since $\gcd(2,3) = 1$, this forces $3 \mid n$: $n$ must be a multiple of $3$, not of $2$. A counterexample to the statement is $n = 3$: $r = 2$ works, the constant term is ${}^6C_2 = 15$, and $3$ is not a multiple of $2$. False.

Question 40

The number of terms in the expansion of $(a+b)^n$ where $n \in \mathbf N$ is one less than the power $n$.

Solution. $(a+b)^n$ has $n + 1$ terms, one more than the power, not one less. False.

Common mistakes

• Question 25: writing ${}^{30}C_{30}$ or $2^{30}$. The largest coefficient is the middle one, ${}^{30}C_{15}$; $2^{30}$ is the sum of all coefficients.
• Question 26: answering $n+1$, the count for a two-variable binomial. Three variables give $\frac{(n+1)(n+2)}{2}$ terms, because each power of $x$ splits into a further binomial in $y$ and $z$.
• Question 28: equating the two bases instead of the exponents and finding no solution. The bases $2^{1/3}$ and $3^{-1/3}$ differ, so the common exponent $n – 12$ itself must be zero.
• Question 30: treating $ba$ as one letter and losing a power of $a$. The term is $(a^3)^{14}(ba)^{14}$, which carries $a^{42+14} = a^{56}$.
• Question 33: writing the remainder as $-1$. A remainder is non-negative, so $-1$ becomes $13 – 1 = 12$.
• Question 34: trusting the printed False. The symmetry argument needs the middle term ${}^{20}C_{10}$ counted once, which is exactly what the formula $2^{19} + \frac{{}^{20}C_{10}}{2}$ does.
• Question 37 and 40: mixing up the number of terms ($n + 1$) with the power ($n$), and the number of terms with the position of the middle term.
• Question 39: assuming the answer is a multiple of $2$ because of the $2n$ in the exponent. The condition comes from dividing $2n$ by $3$, not by $2$.

Practise next

• The rest of this unit: Short Answer Type Q1–11 and Long Answer and Objective Type Q12–24.
• NCERT Class 11 Maths, Binomial Theorem, Exercise 7.1 for remainder and divisibility questions.
• NCERT Class 11 Maths, Binomial Theorem, Miscellaneous Exercise for further coefficient and divisibility problems.