JEE Mathematics Binomial Theorem MCQs Set 10

Answer: (c) \( T_5,T_6 \)

Explanation:
1. Numerical coefficients \( {}^6C_r2^{6-r}5^r \) for \( r=0..6 \): 64, 960, 6000, 20000, 37500, 37500, 15625
2. The greatest value 37500 occurs for \( r=4 \) and \( r=5 \)
3. These are \( T_5 \) and \( T_6 \)

Teacher's Note:
1. Tabulate the numerical coefficient for each r.
2. Two equal greatest terms are possible.

Answer: (b) 2

Explanation:
1. Rearranging: \( {}^{39}C_{3r-1}+{}^{39}C_{3r}={}^{39}C_{r^2}+{}^{39}C_{r^2-1} \), that is \( {}^{40}C_{3r}={}^{40}C_{r^2} \)
2. So \( 3r=r^2 \) or \( 3r+r^2=40 \)
3. \( r=3 \) or \( r=5 \) (the other roots are 0 and -8)
4. Two values

Teacher's Note:
1. Move terms and apply Pascal's rule on both sides.
2. Use nCa = nCb implies a = b or a + b = n.

Answer: (c) \( \frac2{n!}(2^{n-1}-1) \)

Explanation:
1. Multiply and divide by \( n! \): the sum is \( \frac1{n!}\sum_{r=1}^{n-1}{}^nC_r \)
2. \( \sum_{r=1}^{n-1}{}^nC_r=2^n-2 \)
3. So the sum is \( \frac{2^n-2}{n!}=\frac2{n!}(2^{n-1}-1) \)

Teacher's Note:
1. Recognise each term as a binomial coefficient over n factorial.
2. Remove the first and last coefficients from the full sum.

Answer: (c) -2870

Explanation:
1. The product has degree 20, so the coefficient of \( x^{19} \) is minus the sum of the constants
2. \( -(1^2+2^2+\dots+20^2)=-\frac{20\cdot21\cdot41}6=-2870 \)

Teacher's Note:
1. Use the sum of the roots with a negative sign.
2. Apply the sum of squares formula.

Answer: (c) \( \frac12\left[(1+2+\dots+100)^2-(1^2+2^2+\dots+100^2)\right] \)

Explanation:
1. The coefficient of \( x^{98} \) is \( S_2 \), the sum of products of the numbers taken two at a time (signs cancel because \( (-1)^2=1 \))
2. \( S_2=\frac12\left[(\sum k)^2-\sum k^2\right] \)

Teacher's Note:
1. Use the identity (sum) squared = sum of squares + 2 times the sum of pair products.
2. Solve for the pair products.

Answer: (b) \( 2^n \)

Explanation:
1. The two brackets are the real and imaginary parts of \( (1+i)^n \)
2. The expression is \(
3. (1+i)^n
4. ^2=\left(\sqrt2\right)^{2n}=2^n \)

Teacher's Note:
1. Treat the expression as modulus squared of a complex number.
2. The modulus of 1 + i is root 2.

Answer: (b) \( {}^{n+2}C_{r+1} \)

Explanation:
1. \( \left({}^nC_{r+1}+{}^nC_r\right)+\left({}^nC_r+{}^nC_{r-1}\right) \)
2. \( ={}^{n+1}C_{r+1}+{}^{n+1}C_r \)
3. \( ={}^{n+2}C_{r+1} \)

Teacher's Note:
1. Split 2 times nCr into two copies and apply Pascal's rule.
2. Apply it again to combine the results.

Answer: (b) \( \frac12\,{}^{20}C_{10} \)

Explanation:
1. The full alternating sum is 0, and by symmetry the terms from \( {}^{20}C_{11} \) to \( {}^{20}C_{20} \) equal those from \( {}^{20}C_9 \) down to \( {}^{20}C_0 \) with the same sign
2. So \( 2S-{}^{20}C_{10}=0 \)
3. \( S=\frac12\,{}^{20}C_{10} \)

Teacher's Note:
1. Use the symmetry nCr = nC(n - r).
2. The middle term appears only once.

Answer: (a) \( \frac{n-4}5 \)

Explanation:
1. \( T_5={}^nC_4a^{n-4}b^4 \) and \( T_6=-{}^nC_5a^{n-5}b^5 \)
2. \( T_5+T_6=0\Rightarrow{}^nC_4\,a={}^nC_5\,b \)
3. \( \frac ab=\frac{{}^nC_5}{{}^nC_4}=\frac{n-4}5 \)

Teacher's Note:
1. Write the two terms with their signs.
2. Use the ratio of consecutive binomial coefficients.

Answer: (b) Statement-1 is true, Statement-2 is false

Explanation:
1. \( S_1=n(n-1)2^{n-2}=90\cdot2^8 \)
2. \( S_2=n2^{n-1}=10\cdot2^9=20\cdot2^8 \), so the claim \( S_2=10\times2^8 \) is false
3. \( S_3=S_1+S_2=110\cdot2^8=55\cdot2^9 \), so Statement-1 is true

Teacher's Note:
1. Use the standard sums for j and j(j - 1) times C.
2. Then S3 = S1 + S2.

Answer: (c) -144

Explanation:
1. \( 1-x-x^2+x^3=(1-x)(1-x^2)=(1-x)^2(1+x) \)
2. So the expression is \( (1-x)^{12}(1+x)^6 \)
3. Expanding and collecting the coefficient of \( x^7 \) gives \( -144 \)

Teacher's Note:
1. Factorise the polynomial before raising to the power.
2. Combine the two binomial expansions.

Answer: (b) 5th

Explanation:
1. Sum of coefficients \( =3^n=6561\Rightarrow n=8 \)
2. At \( x=\frac12 \), \( (2x)^r=1 \), so the terms equal \( {}^8C_r \)
3. The greatest is \( {}^8C_4 \), the 5th term

Teacher's Note:
1. Put x = 1 to find n.
2. With 2x = 1 the terms are the binomial coefficients themselves.

Answer: (b) 6

Explanation:
1. \( \frac2{a+b}=a-b \) where \( a^2-b^2=2 \)
2. The expression is \( (a+b)^6+(a-b)^6=2\left(a^6+15a^4b^2+15a^2b^4+b^6\right) \)
3. Only even powers of b survive and \( b^2 \), \( a^2 \) are polynomials of degree 1 in \( x^2 \)
4. The highest power is \( x^6 \), so the degree is 6

Teacher's Note:
1. Rationalise the second term to relate it to the first.
2. The sum of conjugate powers is a polynomial.

Answer: (a) \( \sum_{r=0}^5{}^nC_{15-3r}\,{}^nC_r \)

Explanation:
1. \( 1+x+x^3+x^4=(1+x)(1+x^3) \)
2. So \( (1+x)^n(1+x^3)^n \)
3. A term \( {}^nC_rx^{3r} \) from the second factor needs \( x^{15-3r} \) from the first, with coefficient \( {}^nC_{15-3r} \)
4. Sum over \( r=0,\dots,5 \)

Teacher's Note:
1. Factorise the polynomial first.
2. Match powers of x from the two factors.

Answer: (d) \( r\ge n \)

Explanation:
1. \( \frac1{1-x}=\sum x^k \), so the coefficient of \( x^r \) is \( \sum_{j=0}^{\min(r,n)}{}^nC_j \)
2. This equals \( 2^n \) only when all coefficients are included, that is \( r\ge n \)

Teacher's Note:
1. Dividing by 1 - x adds up the coefficients cumulatively.
2. The full sum 2 to the n needs all n + 1 terms.

Answer: (d) \( 4^n-3^n \)

Explanation:
1. \( \sum_{p=0}^{r-1}{}^rC_p2^p=3^r-2^r \)
2. The sum becomes \( \sum_{r=1}^n{}^nC_r(3^r-2^r)=(4^n-1)-(3^n-1) \)
3. \( =4^n-3^n \)

Teacher's Note:
1. Complete the inner sum by adding and removing the last term.
2. Then use the binomial theorem twice.

Answer: (a) \( (-1)^n2^n\left(\frac{3n}2+1\right) \)

Explanation:
1. The sum is \( \sum(r+1)\,{}^nC_r(-3)^r=(1-3)^n+n(-3)(1-3)^{n-1} \)
2. \( =(-2)^n-3n(-2)^{n-1}=(-2)^{n}\left(1+\frac{3n}2\right) \)

Teacher's Note:
1. Split (r + 1) into r and 1.
2. Use the identity r times nCr equals n times (n - 1)C(r - 1).

Answer: (d) 6

Explanation:
1. \( \left(3x^2-\frac1{3x^2}\right)^4 \) has terms \( {}^4C_r(3x^2)^{4-r}\left(-\frac1{3x^2}\right)^r \) with power \( x^{8-4r} \)
2. Powers present: \( x^8,x^4,1,x^{-4},x^{-8} \)
3. Constant term of this factor (r = 2): \( 6\cdot9\cdot\frac19=6 \)
4. The factors 1, x and 2x squared of the first bracket do not combine with these powers to give another constant, so the answer is \( 1\times6=6 \)

Teacher's Note:
1. List which powers of x occur in the second bracket.
2. Only the constant term from the first bracket matches the constant term of the second.

Answer: (c) n = 999 or n = 1000

Explanation:
1. \( \frac{a_{n+1}}{a_n}=\frac{1000}{n+1} \)
2. This is greater than 1 for \( n<999 \) and equals 1 at \( n=999 \)
3. So \( a_{1000}=a_{999} \) and these two are the greatest

Teacher's Note:
1. Compare consecutive terms through their ratio.
2. A ratio of exactly 1 gives two equal greatest terms.

Answer: (a) \( {}^nC_k \)

Explanation:
1. Use \( \binom nj\binom{n-j}{k-j}=\binom nk\binom kj \)
2. The sum becomes \( \binom nk\sum_j(-1)^j\binom kj2^{k-j}=\binom nk(2-1)^k \)
3. \( ={}^nC_k \)

Teacher's Note:
1. Rewrite the product of binomial coefficients in a common form.
2. The remaining sum is a binomial expansion.