JEE Mathematics Progressions Related Inequalities and Series MCQs Set 12

Answer: (c) 6

Explanation:
1. The expression has six terms \( ab^2,ac^2,bc^2,ba^2,ca^2,cb^2 \).
2. Their product is \( (abc)^6 \), so by AM-GM the sum is at least \( 6abc \).
3. Equality at \( a=b=c \), so \( \lambda=6 \).

Teacher's Note:
1. The six terms multiply to a perfect sixth power of abc.

Answer: (c) a rational number that is not an integer

Explanation:
1. \( \int_0^nx\,dx=\tfrac{n^2}2 \).
2. \( \sum_{n=1}^{10}\tfrac{n^2}2=\tfrac12\cdot385=\tfrac{385}2 \).
3. This is rational but not an integer.

Teacher's Note:
1. Evaluate each integral first, then use the sum of squares formula.

Answer: (b) \( \tfrac{2000}{9801} \)

Explanation:
1. The terms are \( \dfrac{2k}{10^{2k-1}} \), i.e. \( \tfrac1{10}\cdot2k\left(\tfrac1{100}\right)^{k-1} \).
2. With \( \sum kx^{k-1}=\dfrac1{(1-x)^2} \) and \( x=\tfrac1{100} \), the sum is \( \tfrac1{10}\cdot2\cdot\dfrac1{(0.99)^2} \).
3. \( =\dfrac{0.2}{0.9801}=\dfrac{2000}{9801} \).

Teacher's Note:
1. This is an arithmetico-geometric series with ratio 1/100.

Answer: (c) \( \tfrac12\left(\sum_{r=1}^nr^2-\sum_{r=1}^nr\right) \)

Explanation:
1. \( \sum_{r=1}^mr=\tfrac{m(m+1)}2 \), so the double sum is \( \tfrac12\sum_{m=1}^n(m^2+m) \).
2. The expression becomes \( \sum r^2-\tfrac12\sum r^2-\tfrac12\sum r \).
3. \( =\tfrac12\left(\sum r^2-\sum r\right) \).

Teacher's Note:
1. Evaluate the inner sum first.

Answer: (a) 255

Explanation:
1. Multiplying by \( (1-x) \) telescopes the product: \( (1-x)\prod=1-x^{256} \).
2. So the product is \( \dfrac{1-x^{256}}{1-x}=1+x+\cdots+x^{255} \).
3. Hence \( n=255 \).

Teacher's Note:
1. Multiply by 1 - x repeatedly using the difference of squares.

Answer: (b) \( -55 \)

Explanation:
1. \( \sum_{i\lt j}x_ix_j=\tfrac12\left[(\sum x_i)^2-\sum x_i^2\right] \).
2. The numbers cancel in pairs so \( \sum x_i=0 \), and \( \sum x_i^2=2(1+4+9+16+25)=110 \).
3. The result is \( \tfrac12(0-110)=-55 \).

Teacher's Note:
1. Use the square of the sum identity for products taken two at a time.

Answer: (d) none of these

Explanation:
1. \( \dfrac1{\log_{2^k}4}=\log_42^k=\dfrac k2 \).
2. The sum is \( \tfrac12\sum_{k=1}^nk=\dfrac{n(n+1)}4 \).
3. This does not match the first three options.

Teacher's Note:
1. Use the reciprocal rule for logarithms to turn each term into k/2.

Answer: (c) 4

Explanation:
1. For three terms in GP: \( \left(\tfrac{\sqrt{10}}3\cdot\tfrac{n(n+1)(2n+1)}6\right)^2=\tfrac{n(n+1)}2\cdot\left(\tfrac{n(n+1)}2\right)^2 \).
2. This simplifies to \( 80(2n+1)^2=324\,n(n+1) \), i.e. \( n^2+n-20=0 \).
3. So \( n=4 \).

Teacher's Note:
1. Use the squares-of-middle-term condition with the standard summation formulas.

Answer: (c) \( n-1+\dfrac1{2^n} \)

Explanation:
1. The numerator is \( 2^k-1 \), so each term is \( 1-2^{-k} \).
2. \( S_n=n-\sum_{k=1}^n2^{-k}=n-\left(1-2^{-n}\right) \).
3. \( S_n=n-1+\dfrac1{2^n} \).

Teacher's Note:
1. Sum the geometric series in the numerator first.

Answer: (a) \( \dfrac{\pi^4}{96} \)

Explanation:
1. Let \( S=\tfrac{\pi^4}{90} \). The even-numbered terms sum to \( \sum\dfrac1{(2r)^4}=\dfrac S{16} \).
2. The odd-numbered terms sum to \( S-\dfrac S{16}=\dfrac{15}{16}S \).
3. \( =\dfrac{15}{16}\cdot\dfrac{\pi^4}{90}=\dfrac{\pi^4}{96} \).

Teacher's Note:
1. Subtract the even-denominator terms from the full sum.

Answer: (b) \( \dfrac{21!-1}{21!} \)

Explanation:
1. \( \dfrac n{(n+1)!}=\dfrac{(n+1)-1}{(n+1)!}=\dfrac1{n!}-\dfrac1{(n+1)!} \).
2. The sum telescopes to \( \dfrac1{1!}-\dfrac1{21!} \).
3. \( =\dfrac{21!-1}{21!} \).

Teacher's Note:
1. Write n as (n + 1) - 1 so the terms telescope.

Answer: (b) \( 49^2+2 \)

Explanation:
1. Successive differences are \( 1,3,5,7,\dots \), the odd numbers.
2. So \( t_n=2+\sum_{k=1}^{n-1}(2k-1)=2+(n-1)^2 \).
3. \( t_{50}=49^2+2 \).

Teacher's Note:
1. The differences of the terms form an AP of odd numbers.

Answer: (b) 2

Explanation:
1. The \( k \)th factor is \( (2^k)^{1/2^{k+1}}=2^{k/2^{k+1}} \).
2. The exponent is \( \sum_{k\ge1}\dfrac k{2^{k+1}}=\tfrac12\sum\dfrac k{2^k}=\tfrac12\cdot2=1 \).
3. The product is \( 2^1=2 \).

Teacher's Note:
1. Collect all the powers of 2 into one exponent and sum it.

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

Explanation:
1. For odd \( n \), the first \( n-1 \) terms form an even-length sum equal to \( \dfrac{(n-1)n^2}2 \).
2. The \( n \)th term is \( n^2 \) (odd positions carry no factor 2).
3. The total is \( \dfrac{(n-1)n^2}2+n^2=\dfrac{n^2(n+1)}2 \).

Teacher's Note:
1. Add the last term to the known sum for the even count n - 1.

Answer: (a) \( \dfrac{(n+1)^2(2n-1)}4 \)

Explanation:
1. For \( n=1 \) the value is 1 and for \( n=3 \) it is \( 27-8+1=20 \); for \( n=5 \) it is \( 125-64+27-8+1=81 \).
2. Formula (a) gives \( \tfrac{4\cdot1}4=1 \), \( \tfrac{16\cdot5}4=20 \), \( \tfrac{36\cdot9}4=81 \).
3. Formulas (b) and (c) fail at \( n=3 \) (giving 5 and 28).

Teacher's Note:
1. Test the candidate formulas on small odd values of n.

Answer: (c) \( (n^2-n+1)+(n^2-n+3)+(n^2-n+5)+\cdots+(n^2+n-1) \)

Explanation:
1. The first term of the row for \( n \) is \( n^2-n+1 \) (1, 3, 7, 13, ...).
2. The \( n \) consecutive odd terms then have average \( n^2 \), since their middle value is \( (n^2-n+1)+(n-1)=n^2 \).
3. So the sum is \( n\cdot n^2=n^3 \) and the last term is \( n^2+n-1 \).

Teacher's Note:
1. The first terms 1, 3, 7, 13 are n squared minus n plus 1.

Answer: (b) \( \dfrac{2^{21}-1}{2^{10}}+19 \)

Explanation:
1. \( t_r^2=2^r+2^{-r}+2 \).
2. \( \sum_{r=1}^{10}2^r=2^{11}-2 \), \( \sum2^{-r}=1-2^{-10} \), and \( \sum2=20 \).
3. Total \( =2^{11}-2^{-10}+19=\dfrac{2^{21}-1}{2^{10}}+19 \).

Teacher's Note:
1. Square t first, then sum the two geometric series.

Answer: (d) \( \dfrac{n(n+3)}2 \)

Explanation:
1. \( S_k=\dfrac1{1-\frac1{k+1}}=\dfrac{k+1}k \), so \( kS_k=k+1 \).
2. \( \sum_{k=1}^n(k+1)=\dfrac{n(n+1)}2+n \).
3. \( =\dfrac{n(n+3)}2 \).

Teacher's Note:
1. Each S_k is an infinite geometric series with ratio 1/(k+1).

Answer: (c) 16! - 1

Explanation:
1. \( n\cdot n!=(n+1)!-n! \).
2. The sum telescopes: \( \sum_{n=1}^{15}\left[(n+1)!-n!\right]=16!-1! \).
3. The value is \( 16!-1 \).

Teacher's Note:
1. Write n as (n + 1) - 1 and multiply by n factorial.

Answer: (b) 51

Explanation:
1. \( f(n)=0 \) while \( \tfrac12+\tfrac n{100}\lt1 \), i.e. \( n\le49 \).
2. For \( 50\le n\le100 \), \( 1\le\tfrac12+\tfrac n{100}\le1.5 \), so \( f(n)=1 \).
3. There are 51 such values of \( n \), so the sum is 51.

Teacher's Note:
1. Find the range of n for which the bracket is 0 and where it is 1.

Answer: (c) \( (\sqrt2)^{n+1} \)

Explanation:
1. The GP has ratio 2, so \( x_n=2^{n-1} \).
2. \( y_n^2=4x_n=2^{n+1} \).
3. In the first quadrant \( y_n\gt0 \), so \( y_n=2^{(n+1)/2}=(\sqrt2)^{n+1} \).

Teacher's Note:
1. Find x_n first and substitute into the parabola equation.