JEE Mathematics Binomial Theorem MCQs Set 20

Answer: (b) \( k \) is a prime number

Explanation:
1. Put \( x=1 \): \( 0=1+a_1+a_2+\dots+a_{12} \). Put \( x=-1 \): \( (1-1-2)^6=64=1-a_1+a_2-a_3+\dots+a_{12} \).
2. Adding the two relations: \( 64=2+2(a_2+a_4+\dots+a_{12}) \), so \( k=31 \).
3. 31 is a prime number, and it is neither a perfect square nor a perfect cube, and it is less than 64.

Teacher's Note:
1. Substituting \( x=1 \) and \( x=-1 \) separates the sums of even-indexed and odd-indexed coefficients.
2. Do not forget the constant term \( a_0=1 \) when adding.

Answer: (a) 0, if \( n \) is odd

Explanation:
1. \( f\left(-\frac1x\right)=1+x \), so the expression is \( (1+x)^n\left(1-\frac1x\right)^n=(-1)^n\frac{(1-x^2)^n}{x^n} \).
2. The term independent of \( x \) is the coefficient of \( x^n \) in \( (-1)^n(1-x^2)^n \). This expansion has only even powers of \( x \).
3. If \( n \) is odd, there is no \( x^n \) term, so the answer is 0. If \( n \) is even, the coefficient is \( (-1)^{n}(-1)^{n/2}\,{}^nC_{n/2}=(-1)^{n/2}\,{}^nC_{n/2} \), which differs from the expression in (c). So only (a) is correct.

Teacher's Note:
1. Simplify the product first: here \( f(-1/x) \) cancels the powers of \( x \) neatly.
2. Check the parity of the exponent before looking for a coefficient.

Answer: (a) \( (n-1)\,{}^{2n}C_n+2^{2n} \)

Explanation:
1. Expand: \( \sum_{i<j}(C_i^2+C_j^2+2C_iC_j)=n\sum_{r=0}^nC_r^2+2\sum_{i<j}C_iC_j \), because each \( C_r^2 \) appears in \( n \) pairs.
2. Use \( \sum C_r^2={}^{2n}C_n \) and \( 2\sum_{i<j}C_iC_j=\left(\sum C_r\right)^2-\sum C_r^2=2^{2n}-{}^{2n}C_n \).
3. The total is \( n\,{}^{2n}C_n+2^{2n}-{}^{2n}C_n=(n-1)\,{}^{2n}C_n+2^{2n} \). Check for \( n=2 \): \( 9+4+9=22=1\cdot6+16 \).

Teacher's Note:
1. Count how many times each \( C_r^2 \) appears when summing over pairs.
2. Always test the final formula with a small value of \( n \).

Answer: (a) 286

Explanation:
1. Put \( x=x'+1,\ y=y'+2,\ z=z'+3,\ w=w'+4 \) with \( x',y',z',w'\ge0 \).
2. Then \( x'+y'+z'+w'=20-10=10 \).
3. The number of non-negative solutions is \( {}^{10+4-1}C_{4-1}={}^{13}C_3=286 \).

Teacher's Note:
1. Shift each variable by its lower bound to reduce to non-negative solutions.
2. The number of non-negative solutions of \( x_1+\dots+x_k=m \) is \( {}^{m+k-1}C_{k-1} \).

Answer: (b) 1

Explanation:
1. Let \( F=(2-\sqrt3)^n \). Since \( 0<2-\sqrt3<1 \), we have \( 0<F<1 \).
2. \( I+f+F=(2+\sqrt3)^n+(2-\sqrt3)^n=2\left[{}^nC_02^n+{}^nC_22^{n-2}\cdot3+\dots\right] \), an integer. Hence \( f+F \) is an integer; as \( 0<f+F<2 \), \( f+F=1 \), i.e. \( F=1-f \).
3. \( (I+f)(1-f)=(2+\sqrt3)^n(2-\sqrt3)^n=1^n=1 \).

Teacher's Note:
1. Pair \( (2+\sqrt3)^n \) with its conjugate: the sum is an integer and the conjugate is a proper fraction.
2. \( (2+\sqrt3)(2-\sqrt3)=1 \).

Answer: (c) \( 3r+2 \)

Explanation:
1. \( (1+x)^{101}(1-x+x^2)^{100}=(1+x)\left[(1+x)(1-x+x^2)\right]^{100}=(1+x)(1+x^3)^{100} \).
2. \( (1+x^3)^{100}=\sum{}^{100}C_rx^{3r} \), so the product is \( \sum{}^{100}C_rx^{3r}+\sum{}^{100}C_rx^{3r+1} \).
3. Only powers \( 3r \) and \( 3r+1 \) occur, so there is no term in \( x^{3r+2} \).

Teacher's Note:
1. Use \( (1+x)(1-x+x^2)=1+x^3 \) to collapse the product.
2. Powers of the form \( 3r+2 \) never appear.

Answer: (b) (ii) and (iii) only

Explanation:
1. Replacing each index \( i \) by \( n-i+1 \) permutes the distinct triples among themselves, so \( S_n=\sum\sum\sum(-x_{n-i+1})(-x_{n-j+1})(-x_{n-k+1}) \).
2. The product of three sign changes gives \( -x_ix_jx_k \) overall, so \( S_n=-S_n \) and \( S_n=0 \) for every \( n \).
3. Hence \( S_{10}=0\ne121 \); \( S_{10}=S_{20}=0 \); \( S_{14}=0 \); and \( S_{30}=S_{31}=0 \), so \( S_{30}>S_{31} \) is false. Statements (ii) and (iii) are correct.

Teacher's Note:
1. A symmetry argument showing \( S=-S \) is much faster than computing the sum.
2. Check each statement separately once you know the value is 0.

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. \( T_{r+1}={}^{256}C_r\,3^{(256-r)/2}\,5^{r/8} \). The term is an integer when \( r \) is a multiple of 8, i.e. \( r=0,8,16,\dots,256 \). That gives \( 33 \) terms, so \( k=33 \): (i) and (ii) are true.
2. The number of distinct terms in \( (x+y-z)^{16} \) is the number of non-negative solutions of \( a+b+c=16 \), namely \( {}^{18}C_2=153 \).
3. So \( 153>140 \) is true and \( 153<150 \) is false: (iii) true, (iv) false.

Teacher's Note:
1. For a term to be an integer, both exponents \( \frac{256-r}2 \) and \( \frac r8 \) must be integers.
2. The number of distinct terms of a trinomial power \( n \) is \( {}^{n+2}C_2 \).

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. Use \( n\,{}^{n-1}C_{r-1}=r\,{}^nC_r \). The bracket becomes \( r^2\,{}^nC_r-r^2\,{}^nC_{r-1}+(2r+1)\,{}^nC_r=(r+1)^2\,{}^nC_r-r^2\,{}^nC_{r-1} \).
2. The sum telescopes: \( f(n)=(n+1)^2\,{}^nC_n-1^2\,{}^nC_0=(n+1)^2-1 \).
3. \( f(10)=121-1=120 \) and \( f(20)=441-1=440 \). \( \sum_{n=1}^{10}f(n)=\left(2^2+3^2+\dots+11^2\right)-10=505-10=495 \). So (i), (ii), (iii) are correct and (iv) is wrong.

Teacher's Note:
1. Rewrite each term as a difference \( g(r)-g(r-1) \) so the sum telescopes.
2. Check: \( f(1)=3 \) directly from the definition equals \( 2^2-1 \).

Answer: (a) (i) and (iii) only

Explanation:
1. Let \( (\sqrt3+1)^{2m}=I+F \) with \( 0<F<1 \), and \( G=(\sqrt3-1)^{2m} \) with \( 0<G<1 \).
2. \( I+F+G=(4+2\sqrt3)^m+(4-2\sqrt3)^m=2^m\left[(2+\sqrt3)^m+(2-\sqrt3)^m\right]=2^{m+1}\times\text{integer} \). So \( F+G \) is an integer between 0 and 2, hence \( F+G=1 \).
3. Then \( I+1=2^{m+1}\times\text{integer} \): the integer just greater than \( (\sqrt3+1)^{2m} \) is divisible by \( 2^{m+1} \) and therefore by \( 2^m \). Divisibility by powers of 3 fails (for \( m=1 \) the number is 8).

Teacher's Note:
1. Pair the number with its conjugate to make an integer sum.
2. \( (\sqrt3+1)^2=2(2+\sqrt3) \) gives the factor \( 2^m \).

Answer: (a) 5

Explanation:
1. \( \frac{r+2}{r+1}=1+\frac1{r+1} \), and \( \frac{{}^nC_r}{r+1}=\frac{{}^{n+1}C_{r+1}}{n+1} \).
2. So the sum is \( 2^n+\frac1{n+1}\left(2^{n+1}-1\right) \).
3. For \( n=5 \): \( 32+\frac{63}6=\frac{255}6=\frac{2^8-1}6 \), so \( n=5 \).

Teacher's Note:
1. Split the fraction and use \( \frac{{}^nC_r}{r+1}=\frac{{}^{n+1}C_{r+1}}{n+1} \).
2. The left side increases with \( n \), so one value of \( n \) fits.

Answer: (a) 6

Explanation:
1. \( a_1=n-2 \) and \( a_2=\frac{n(n-1)}2-2n+5 \).
2. \( a_1^2=2a_2 \Rightarrow (n-2)^2=n(n-1)-4n+10 \), that is \( n^2-4n+4=n^2-5n+10 \).
3. Hence \( n=6 \).

Teacher's Note:
1. Only the first three coefficients of the product are needed.
2. Compare \( x \) and \( x^2 \) terms from each combination.

Answer: (a) 2

Explanation:
1. Put \( x=1,\omega,\omega^2 \) (cube roots of unity) and add: \( 3(a_0+a_3+a_6+\dots)=3^n \), so \( a_0+a_3+a_6+\dots=3^{n-1} \).
2. Multiply the expansion by \( x^2 \) and repeat: \( a_1+a_4+a_7+\dots=3^{n-1} \), and likewise \( a_2+a_5+a_8+\dots=3^{n-1} \).
3. The required ratio is \( \frac{3^{n-1}+3^{n-1}}{3^{n-1}}=2 \).

Teacher's Note:
1. Roots of unity filter coefficients by the residue of the exponent modulo 3.
2. Each residue class has the same sum \( 3^{n-1} \) here.

Answer: (a) 3

Explanation:
1. Comparing coefficients: \( na=8 \) and \( \frac{n(n-1)}2a^2=24 \).
2. From the first, \( n=\frac8a \). Substituting: \( \left(\frac8a\right)\left(\frac8a-1\right)a^2=48\Rightarrow64-8a=48\Rightarrow a=2 \), so \( n=4 \).
3. \( 9\cdot\frac{4-2}{2+4}=9\cdot\frac13=3 \).

Teacher's Note:
1. Match coefficients of \( x \) and \( x^2 \) to find both unknowns.
2. Substitute \( n=8/a \) to get one equation in \( a \).

Answer: (c) \( na \)

Explanation:
1. Use \( r\,{}^nC_r=n\,{}^{n-1}C_{r-1} \): the sum becomes \( na\sum_{r=1}^n{}^{n-1}C_{r-1}a^{r-1}b^{n-r} \).
2. The remaining sum is \( (a+b)^{n-1} \).
3. So the value is \( na(a+b)^{n-1}=na \), since \( a+b=1 \).

Teacher's Note:
1. The identity \( r\,{}^nC_r=n\,{}^{n-1}C_{r-1} \) lowers the degree of the binomial.
2. This is the mean of a binomial distribution \( B(n,a) \).

Answer: (c) \( n^2p^2+npq \)

Explanation:
1. Write \( r^2=r(r-1)+r \). Then \( \sum r(r-1)\,{}^nC_rp^rq^{n-r}=n(n-1)p^2(p+q)^{n-2}=n(n-1)p^2 \).
2. And \( \sum r\,{}^nC_rp^rq^{n-r}=np(p+q)^{n-1}=np \).
3. The total is \( n(n-1)p^2+np=n^2p^2-np^2+np=n^2p^2+np(1-p)=n^2p^2+npq \).

Teacher's Note:
1. Splitting \( r^2=r(r-1)+r \) is the standard technique for such sums.
2. This is \( E(X^2)=\operatorname{Var}(X)+[E(X)]^2 \) for a binomial distribution.

Answer: (c) \( n(n-1)2^{n-3} \)

Explanation:
1. For a fixed \( j \), \( i \) runs over \( 0,1,\dots,j-1 \), whose sum is \( \frac{j(j-1)}2 \). So the sum equals \( \frac12\sum_{j}j(j-1)\,{}^nC_j \).
2. \( \sum j(j-1)\,{}^nC_j=n(n-1)2^{n-2} \).
3. The value is \( \frac12n(n-1)2^{n-2}=n(n-1)2^{n-3} \). Check for \( n=3 \): \( 3+3=6=3\cdot2\cdot1 \).

Teacher's Note:
1. Sum over \( i \) first when the summand has \( i \) as a factor and only \( j \) in the binomial coefficient.
2. \( \sum j(j-1)\,{}^nC_j=n(n-1)2^{n-2} \) is a standard result.

Answer: (a) \( n(3n+1)2^{n-3} \)

Explanation:
1. For fixed \( i \), \( j \) runs over \( i+1,\dots,n \), with sum \( \frac{(n-i)(n+i+1)}2 \). So the sum is \( \frac12\sum_{i=0}^{n}{}^nC_i(n-i)(n+i+1) \).
2. Expand \( (n-i)(n+i+1)=n(n+1)-i-i^2 \). Using \( \sum{}^nC_i=2^n \), \( \sum i\,{}^nC_i=n2^{n-1} \), \( \sum i^2\,{}^nC_i=n(n+1)2^{n-2} \), the sum is \( \frac12\left[n(n+1)2^n-n2^{n-1}-n(n+1)2^{n-2}\right] \).
3. Simplify: \( \frac{2^{n-2}}2\,n\left[4(n+1)-2-(n+1)\right]=n(3n+1)2^{n-3} \). Check \( n=3 \): the sum is \( 1+8+21=30=3\cdot10 \).

Teacher's Note:
1. Sum over \( j \) first because \( j \) appears only as a factor.
2. A small case such as \( n=3 \) quickly tests the final formula.

Answer: (a) 5

Explanation:
1. \( T_{r+1}=\frac{\frac72\left(\frac72-1\right)\cdots\left(\frac72-r+1\right)}{r!}x^r \).
2. The coefficient first becomes negative when a factor \( \frac72-r+1 \) becomes negative, i.e. \( r>\frac92 \).
3. So the smallest such \( r \) is 5.

Teacher's Note:
1. For fractional index the series does not terminate and the signs can change.
2. Find the first factor \( \frac72-k \) that is negative.

Answer: (a) 10

Explanation:
1. The general term is \( T_{r+1}={}^5C_r(y^2)^{5-r}\left(\frac1y\right)^r={}^5C_ry^{10-3r} \).
2. For \( y^1 \): \( 10-3r=1\Rightarrow r=3 \).
3. The coefficient is \( {}^5C_3=10 \).

Teacher's Note:
1. Write the general term and equate the exponent to the required power.
2. \( {}^5C_3={}^5C_2=10 \).

Answer: (a) 14

Explanation:
1. Simplify the second part: \( \frac a{\sqrt{a^{-1}}}=a\cdot a^{1/2}=a^{3/2} \).
2. \( T_2={}^nC_1\left(a^{1/13}\right)^{n-1}a^{3/2}=n\,a^{\frac{n-1}{13}+\frac32} \).
3. Matching with \( 14a^{5/2} \): \( n=14 \) and \( \frac{n-1}{13}+\frac32=\frac52 \), which holds for \( n=14 \).

Teacher's Note:
1. Simplify the radicals into a single power of \( a \) before using the general term.
2. Both the coefficient and the power of \( a \) must match.

Answer: (a) 4

Explanation:
1. \( 1+2x+3x^2+\dots=(1-x)^{-2} \).
2. So the expression is \( \left[(1-x)^{-2}\right]^{1/2}=(1-x)^{-1}=1+x+x^2+x^3+\dots \).
3. Each of the coefficients of \( x^2,x^4,x^6,x^8 \) is 1, so their sum is 4.

Teacher's Note:
1. Recognise \( 1+2x+3x^2+\dots \) as \( (1-x)^{-2} \).
2. Taking the square root simplifies it to a geometric series.

Answer: (a) 2

Explanation:
1. \( 99^n+1=(100-1)^n+1=1+\left({}^nC_0100^n-{}^nC_1100^{n-1}+\dots+{}^nC_{n-1}100-{}^nC_n\right) \).
2. For odd \( n \) the constant terms cancel (\( 1-1=0 \)), leaving \( 100\left\{{}^nC_0100^{n-1}-\dots+{}^nC_{n-1}\right\} \) with the last term \( {}^nC_{n-1}=n \).
3. The bracket ends in the digit of \( n \), which is odd and non-zero, so there are exactly two zeros at the end.

Teacher's Note:
1. Write \( 99=100-1 \) so that all terms but the last two are multiples of 100.
2. The unit digit of the bracket decides whether more zeros appear.

Answer: (a) 9

Explanation:
1. Write \( f(n)=(10^n-1)+3\cdot4^{n+2}+6=(10^n-1)+6\left(2^{2n+3}+1\right) \).
2. \( 10^n-1 \) is divisible by 9, and \( 2^{2n+3}+1 \) is divisible by \( 2+1=3 \) (odd power), so \( 6(2^{2n+3}+1) \) is divisible by 18. Hence \( f(n) \) is divisible by 9.
3. \( f(1)=207=9\cdot23 \) and \( f(2)=873=9\cdot97 \), whose greatest common divisor is 9. So the greatest such integer is 9.

Teacher's Note:
1. Rewrite the expression to expose factors such as \( 10^n-1 \) and \( a^{odd}+1 \).
2. Confirm using the first two values and their gcd.

Answer: (a) 6

Explanation:
1. \( x+\frac1x=1\Rightarrow x^2-x+1=0\Rightarrow x=-\omega \) or \( -\omega^2 \), so \( x^6=1 \).
2. \( x^{1000}=x^{4}=x^3\cdot x=-x \), so \( p=-\left(x+\frac1x\right)=-1 \).
3. \( 2^{4n}=16^n \) ends in 6, so \( 2^{4n}+1 \) ends in 7 and \( q=7 \). Therefore \( p+q=-1+7=6 \).

Teacher's Note:
1. Complex roots of \( x^2-x+1=0 \) are sixth roots of unity, so reduce the exponent modulo 6.
2. Powers of 16 always end in 6.

Answer: (a) 0

Explanation:
1. \( 5!,6!,\dots,100! \) all end in 0. So the unit digit of \( \sum_{r=0}^{100}r! \) is that of \( 0!+1!+2!+3!+4!=34 \), namely 4.
2. For even \( n \), \( 2^{2n}=4^n=16^{n/2} \) ends in 6.
3. The unit digit of the total is that of \( 4+6=10 \), namely 0.

Teacher's Note:
1. Factorials from \( 5! \) onwards end in 0, so only the first five matter.
2. Powers of 4 end alternately in 4 and 6; even exponents give 6.