JEE Mathematics Binomial Theorem MCQs Set 07

Answer: (d) \( n^2+n+1 \)

Explanation:
1. The base equals \( 1+x^{-3}+x^3 \). With N = n(n+1)/2, the exponents of x take the values \( -3N,-3N+3,\dots,3N \).
2. That is \( 2N+1=n(n+1)+1=n^2+n+1 \) terms.

Teacher's Note:
1. Rewrite the base as x⁻³ + 1 + x³ (a symmetric trinomial).
2. Count the distinct powers of x³.

Answer: (c) 7

Explanation:
1. \( 5^5=3125=284\cdot11+1 \), so \( 5^{40}=(5^5)^8\equiv1\pmod{11} \) and α = 1.
2. \( 2^8=256=15\cdot17+1 \), and \( 2003=8\cdot250+3 \), so \( 2^{2003}\equiv2^3=8\pmod{17} \): β = 8.
3. β − α = 7.

Teacher's Note:
1. Find a power of the base that is 1 modulo the divisor.
2. Reduce the exponent modulo that power.

Answer: (d) none of these

Explanation:
1. The general term is \( \binom{10}r\left(\frac x3\right)^{\frac{10-r}2}\left(\frac3{2x^2}\right)^{\frac r2} \), with power of x equal to \( \frac{10-r}2-r \).
2. Setting this to 0 gives \( r=\frac{10}3 \), which is not an integer, so there is no term independent of x (the term is 0).

Teacher's Note:
1. Find r from the exponent of x.
2. If r is not a whole number there is no such term.

Answer: (b) \( \frac{(2n+2)!}{\{(n+1)!\}^2} \)

Explanation:
1. For an even index 2n + 2 the greatest binomial coefficient is the middle one: \( \binom{2n+2}{n+1}=\frac{(2n+2)!}{\{(n+1)!\}^2} \).

Teacher's Note:
1. The middle term of an even-index expansion has the greatest coefficient.
2. Write C(2n+2, n+1) in factorials.

Answer: (c) 4th term

Explanation:
1. The general term is \( T_{r+1}=\binom9r3^{\frac{9-r}2}2^{\frac r3} \).
2. Integral requires r to be a multiple of 3 and 9 − r even: r = 0 fails (9 odd); r = 3 works (9 − 3 = 6).
3. So T₄ is the first integral term.

Teacher's Note:
1. The powers 3^((9−r)/2) and 2^(r/3) must both be integers.
2. Test r = 0, 3, 6, 9 in order.

Answer: (c) 50

Explanation:
1. The general term is \( \binom{55}r2^{\frac{55-r}5}3^{\frac r{10}} \).
2. It is rational when r is a multiple of 10 (and then 55 − r is a multiple of 5 too): r = 0, 10, 20, 30, 40, 50 — 6 terms.
3. There are 56 terms in all, so 56 − 6 = 50 are irrational.

Teacher's Note:
1. Both exponents must be integers for a term to be rational.
2. Count the total terms (n + 1) and subtract.

Answer: (c) 120

Explanation:
1. Take \( (bc)^p(ca)^q(ab)^r \) with p + q + r = 10. The powers: a: q + r = 10, b: p + r = 7, c: p + q = 3.
2. Solving: p = 0, q = 3, r = 7.
3. Coefficient \( \frac{10!}{0!\,3!\,7!}=120 \).

Teacher's Note:
1. Match the exponents of a, b and c.
2. Use the multinomial coefficient.

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

Explanation:
1. Use the cube roots of unity: \( \frac13\left[f(1)+f(\omega)+f(\omega^2)\right] \) picks out the coefficients of x⁰, x³, x⁶, ...
2. f(1) = 3ⁿ and f(ω) = f(ω²) = 0 since 1 + ω + ω² = 0.
3. So the sum is \( \frac{3^n}{3}=3^{n-1} \).

Teacher's Note:
1. Roots-of-unity filter for exponents divisible by 3.
2. 1 + ω + ω² = 0.

Answer: (a) 5(2n − 9)

Explanation:
1. \( \frac{^nC_r}{^nC_{r-1}}=\frac{n-r+1}{r} \), so each term is \( n-r+1 \).
2. Sum \( =10(n+1)-\frac{10\cdot11}2=10n-45=5(2n-9) \).

Teacher's Note:
1. Simplify the ratio of consecutive binomial coefficients first.
2. Sum the arithmetic progression.

Answer: (c) A is true, but R is false.

Explanation:
1. \( \frac{{}^nC_r}{r+1}=\frac{{}^{n+1}C_{r+1}}{n+1} \), so \( \sum\frac{{}^nC_rx^r}{r+1}=\frac{1}{(n+1)x}\sum_{r=0}^n{}^{n+1}C_{r+1}x^{r+1}=\frac{(1+x)^{n+1}-1}{(n+1)x} \): A is true.
2. Putting x = 1 gives \( \sum\frac{{}^nC_r}{r+1}=\frac{2^{n+1}-1}{n+1} \), not \( \frac{2^{n+1}}{n+1} \): R is false.

Teacher's Note:
1. Use the identity C(n, r)/(r + 1) = C(n + 1, r + 1)/(n + 1).
2. Put x = 1 to test R.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R is true: for odd n, \( a^n+b^n \) has the factor a + b.
2. Pair k⁹⁹ with (101 − k)⁹⁹ to get multiples of 101, so the sum is divisible by 101. Pair k⁹⁹ with (100 − k)⁹⁹ for k = 1..49; the leftover terms 50⁹⁹ and 100⁹⁹ are also multiples of 100, so the sum is divisible by 100.
3. As gcd(100, 101) = 1, the sum is divisible by 100 × 101 = 10100. A is true, and R is the tool used.

Teacher's Note:
1. Use pairing with R for the moduli 101 and 100 separately.
2. The numbers 100 and 101 are coprime.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. \( \frac{T_{r+1}}{T_r}=\frac{13-r}{r}\cdot\frac{11}{10}\ge1\iff r\le\frac{143}{21}\approx6.8 \), so the greatest term is T₇: A is true.
2. R is also true as a fact, but the greatest term depends on x as well as on the coefficient, so R by itself does not explain A.

Teacher's Note:
1. Compare consecutive terms with the ratio test.
2. The binomial coefficient alone does not decide the greatest term.