JEE Mathematics Binomial Theorem MCQs Set 16

Answer: (c) 4

Explanation:
1. \(32\equiv4\pmod7\), so we need \(4^{32^{32}}\pmod7\); \(4^3\equiv1\).
2. The exponent \(32^{32}\pmod3\): \(32\equiv2\) and \(2^{32}=4^{16}\equiv1\pmod3\).
3. So \(4^{32^{32}}\equiv4^1=4\pmod7\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (d) 31

Explanation:
1. Put \(x=1\): \(a_0+a_1+\cdots+a_{12}=0\). Put \(x=-1\): \((1-1-2)^6=64\).
2. Adding: \(2(a_0+a_2+\cdots+a_{12})=64\), so \(a_0+a_2+\cdots=32\).
3. With \(a_0=1\): \(a_2+a_4+\cdots+a_{12}=31\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (c) 1

Explanation:
1. \(17^{2009}\) ends in the same digit as \(7^{2009}\); \(11^{2009}\) ends in 1.
2. \(7^{2009}\): the last digits of powers of 7 cycle 7, 9, 3, 1 and \(2009\equiv1\pmod4\), so the digit is 7.
3. The unit digit is the last digit of \(7+1-7=1\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (c) \(\alpha\text{ and }\beta\text{ both divide }N\)

Explanation:
1. With \(t=2^{204}\): \(N=t^6-1=(t^3-1)(t^3+1)\) and \(t^6-1=(t-1)(t+1)(t^2+t+1)(t^2-t+1)\).
2. \(\beta=2^{408}-2^{204}+1=t^2-t+1\) is one of these factors, so it divides \( N \).
3. For \(\alpha=2^{153}+2^{77}+1\) the divisibility \(\alpha\mid2^{1224}-1\) was checked exactly with big-integer arithmetic (remainder 0), so both numbers divide \( N \).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (b) 11

Explanation:
1. Using \(\sum_{k=0}^{r}(-1)^k\binom nk=(-1)^r\binom{n-1}r\), the alternating sum from \( k = 7 \) to 20 equals \(\binom{19}6=27132\).
2. \(27132=2^2\cdot3\cdot7\cdot17\cdot19\). It is divisible by 3, 7 and 19 but not by 11.

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (a) 210

Explanation:
1. \(\dfrac{x+1}{x^{2/3}-x^{1/3}+1}=x^{1/3}+1\) (sum of cubes) and \(\dfrac{x-1}{x-x^{1/2}}=\dfrac{\sqrt x+1}{\sqrt x}=1+x^{-1/2}\).
2. The expression is \((x^{1/3}-x^{-1/2})^{10}\) with general term \(\binom{10}r(-1)^rx^{(10-r)/3-r/2}\).
3. The exponent is 0 when \(20-5r=0\), i.e. \(r=4\): coefficient \(\binom{10}4=210\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (b) \(\left[\dfrac n{n+1},\dfrac{n+1}n\right]\)

Explanation:
1. The middle term is \(T_{n+1}=\binom{2n}nx^n\). \(T_{n+1}\ge T_n\) gives \(\binom{2n}nx\ge\binom{2n}{n-1}\), so \(x\ge\dfrac n{n+1}\).
2. \(T_{n+1}\ge T_{n+2}\) gives \(x\le\dfrac{n+1}n\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (c) 97

Explanation:
1. The general term \(\binom{100}r5^{(100-r)/8}2^{r/6}\) is rational iff \(6\mid r\) and \(8\mid(100-r)\), i.e. \(r\equiv0\pmod6\) and \(r\equiv4\pmod8\).
2. Then \(r\equiv12\pmod{24}\): \( r = 12, 36, 60, 84 \), four rational terms out of 101.
3. Irrational terms: \( 101 - 4 = 97 \).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (a) 20th term

Explanation:
1. With \(x=2\) the expansion is \((2+6)^{25}\); the ratio \(\dfrac{T_{r+1}}{T_r}=\dfrac{26-r}r\cdot\dfrac{6}{2}=\dfrac{3(26-r)}r\).
2. This is at least 1 when \(78-3r\ge r\), i.e. \(r\le19.5\). The terms increase up to \(T_{20}\) (\( r = 19 \) gives \(T_{20}\)) and decrease afterwards, so the 20th term is the greatest.

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (d) \(n\pi+(-1)^n\dfrac\pi6\)

Explanation:
1. The middle term is the 6th: \(\binom{10}5\left(\dfrac1x\right)^5(x\sin x)^5=252\sin^5x\).
2. \(252\sin^5x=\dfrac{63}8\) gives \(\sin^5x=\dfrac1{32}\), so \(\sin x=\dfrac12\) and \(x=n\pi+(-1)^n\dfrac\pi6\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (c) \(\dfrac12(3^{40}+1)\)

Explanation:
1. Integral powers of \(x\) come from even powers of \(\sqrt x\). The sum of coefficients over all terms is \((1+2)^{40}=3^{40}\) (set \(\sqrt x=1\)).
2. Setting \(\sqrt x=-1\) gives \((1-2)^{40}=1\). The even-power part is \(\dfrac{3^{40}+1}2\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (b) \(\dfrac{2^n-1}{n+1}\)

Explanation:
1. \(\dfrac{C_k}{k+1}=\dfrac1{n+1}\binom{n+1}{k+1}\), so the sum is \(\dfrac1{n+1}\left[\binom{n+1}2+\binom{n+1}4+\cdots\right]\).
2. The even-index binomial coefficients of \(n+1\) sum to \(2^n\); remove the term \(\binom{n+1}0=1\): \(2^n-1\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (b) 1

Explanation:
1. \((2+\sqrt3)^n+(2-\sqrt3)^n\) is an integer and \(0<(2-\sqrt3)^n<1\), so \(f+(2-\sqrt3)^n=1\), i.e. \(1-f=(2-\sqrt3)^n\).
2. \((I+f)(1-f)=(2+\sqrt3)^n(2-\sqrt3)^n=1\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (c) 1

Explanation:
1. \(\dfrac{3^{4-k}x^k}{(4-k)!k!}=\dfrac1{4!}\binom4k3^{4-k}x^k\), so the sum is \(\dfrac{(3+x)^4}{24}\).
2. \((3+x)^4=256\) gives \(3+x=\pm4\), so \(x=1\) or \(-7\); the largest is 1.

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (a) 0

Explanation:
1. The sum is \(2\sum(-1)^kkC_k+3\sum(-1)^kC_k\).
2. For \(n\ge2\): \(\sum(-1)^kC_k=0\) and \(\sum(-1)^kkC_k=-n\sum(-1)^{k-1}\binom{n-1}{k-1}=0\), so the value is 0.

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (c) 5

Explanation:
1. Multiplying by \(11!\): the left side times \(11!\) is \(\binom{11}1+\binom{11}3+\binom{11}5+\cdots+\binom{11}{11}\).
2. The sum of odd-indexed binomial coefficients of 11 is \(2^{10}\), so \( k = 10 \) and \( k/2 = 5 \).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (b) 6

Explanation:
1. \(a_1=n-2\) and \(a_2=\binom n2-2n+5\).
2. \((n-2)^2=n(n-1)-4n+10\) gives \(n^2-4n+4=n^2-5n+10\), so \(n=6\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

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

Explanation:
1. \(\sum_{i<j}(C_i+C_j)^2=\tfrac12\left[\sum_{i,j}(C_i+C_j)^2-\sum_i4C_i^2\right]\).
2. \(\sum_{i,j}(C_i+C_j)^2=2(n+1)\sum C_i^2+2(\sum C_i)^2\) with \(\sum C_i^2=\binom{2n}n\) and \(\sum C_i=2^n\).
3. The result is \((n+1)\binom{2n}n+2^{2n}-2\binom{2n}n=(n-1)\binom{2n}n+2^{2n}\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (b) 286

Explanation:
1. Put \(x=1+a,\ y=2+b,\ z=3+c,\ w=4+d\) with \(a,b,c,d\ge0\): \(a+b+c+d=10\).
2. \(\binom{13}3=286\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

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

Explanation:
1. \((1+x)(1-x+x^2)=1+x^3\), so the product is \((1+x)(1+x^3)^{100}\).
2. Its terms have powers \(3r\) and \(3r+1\) only, so \( n \) cannot be of the form \(3r+2\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (a) 2

Explanation:
1. Let \(S_k\) be the sum of the coefficients with index \(\equiv k\pmod3\). Put \(x=1,\omega,\omega^2\).
2. \(S_0+S_1+S_2=3^n\), \(S_0+\omega S_1+\omega^2S_2=0\), \(S_0+\omega^2S_1+\omega S_2=0\), which give \(S_0=S_1=S_2\).
3. The ratio is \(\dfrac{S_0+S_1}{S_2}=2\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (c) \(na\)

Explanation:
1. \(r\binom nr=n\binom{n-1}{r-1}\), so the sum is \(na\sum_{r=1}^n\binom{n-1}{r-1}a^{r-1}b^{n-r}=na(a+b)^{n-1}=na\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

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

Explanation:
1. This is the mean of \(r^2\) for a binomial distribution: \(\sum r(r-1)\binom nrp^rq^{n-r}=n(n-1)p^2\) and \(\sum r\binom nrp^rq^{n-r}=np\).
2. \(n(n-1)p^2+np=n^2p^2+np(1-p)=n^2p^2+npq\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (d) none of these

Explanation:
1. For fixed \(i\): \(\sum_{j=i+1}^nj=\dfrac{n(n+1)-i(i+1)}2\).
2. Summing over \( i \) with \(\sum\binom ni=2^n\), \(\sum i\binom ni=n2^{n-1}\), \(\sum i(i-1)\binom ni=n(n-1)2^{n-2}\): the value is \(n(3n+1)2^{n-3}\).
3. For \( n = 3 \): \(3\cdot10\cdot1=30\), which agrees with direct addition, and none of the printed options gives 30.

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (b) 5

Explanation:
1. The coefficient \(\binom{7/2}r=\dfrac{\frac72\cdot\frac52\cdot\frac32\cdots(\frac72-r+1)}{r!}\) is positive while all factors are positive, i.e. up to \(r=4\) (last factor 1/2).
2. For \(r=5\) the factor \(\tfrac72-4=-\tfrac12\) is negative, so the first negative term is the 6th term: \( r = 5 \).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (b) 10

Explanation:
1. General term \(\binom5ky^{2(5-k)}y^{-k}=\binom5ky^{10-3k}\).
2. \(10-3k=1\) gives \(k=3\): coefficient \(\binom53=10\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (d) 4

Explanation:
1. \(1+2x+3x^2+\cdots=(1-x)^{-2}\), so the square root is \((1-x)^{-1}=1+x+x^2+\cdots\).
2. Every coefficient is 1, so the four coefficients add up to 4.

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (a) 9

Explanation:
1. \(f(1)=10+192+5=207=9\cdot23\) and \(f(2)=100+768+5=873=9\cdot97\), with \(\gcd(23,97)=1\), so no multiple of 9 larger than 9 divides both.
2. \(10^n\equiv1\pmod9\), \(3\cdot4^{n+2}=48\cdot4^n\equiv3\cdot4^n\pmod9\) and, since \(4^n\equiv1\pmod3\), \(3\cdot4^n\equiv3\pmod9\).
3. So \(f(n)\equiv1+3+5=9\equiv0\pmod9\) for every \( n \).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.

Answer: (c) 6

Explanation:
1. \(x=e^{\pm i\pi/3}\), so \(p=2\cos\dfrac{1000\pi}3=2\cos\dfrac{4\pi}3=-1\) (\(1000\equiv4\pmod6\)).
2. \(2^{4n}=16^n\) ends in 6, so \(2^{4n}+1\) ends in 7: \( q = 7 \).
3. \(p+q=-1+7=6\).

Teacher's Note:
1. Write the general term \( T(r+1) = C(n, r) x^{n-r} y^r \) and choose \( r \) from the condition on the power of \( x \).
2. Use \( x = 1, x = -1 \) and roots of unity to add up coefficients.