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Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

The sum of the series
$ _{  }^{ 4n }{ { C } _{ 0 } }+ _{  }^{ 4n }{ { C } _{ 4 } }+ _{  }^{ 4n }{ { C } _{ 8 } }+........ _{  }^{ 4n }{ { C } _{ 4n } }$ is 

  1. $2^{4n-2}+(-1)^{n}2^{2n-1}$
  2. $2^{4n-2}+(-1)^{n+1}2^{2n-1}$
  3. $2^{4n-2}-2^{2n-1}$
  4. $2^{4n-2}+2^{2n-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a standard identity for the sum of binomial coefficients with step 4. The sum of C(4n, 4k) is given by (1/4) * [(1+1)^4n + (1-1)^4n + 2 * (1+i)^4n + 2 * (1-i)^4n]. This simplifies to 2^(4n-2) + 2^(2n-1) * cos(n * pi).

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

$1+6+9(\dfrac{1^2 +2^2 +3^2}{7}) +12(\dfrac{1^2 +2^2 +3^2+4^2}{9} )+15(\dfrac{1^2 +2^2 +3^2+4^2+5^2}{11}) +$_____
Find sum of $15$ terms

  1. $7720$
  2. $7820$
  3. $7980$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The general term of the series can be analyzed by looking at the fractions containing sums of squares. The sum of squares of first k integers is k(k+1)(2k+1)/6. Simplifying the denominators and coefficients yields a telescoping or easily summable series whose sum for 15 terms evaluates to 7820.

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

Find the sum of 1 + $\dfrac{1}{4} + \dfrac{1.3}{4.8} + \dfrac{1.3.5}{4.8.12} +.......\infty $

  1. $2\sqrt{2}$
  2. $\sqrt{2}$
  3. $\sqrt{2}$
  4. $\sqrt { \frac { 1 }{ 2 } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is a binomial expansion (1-x)^(-n). The series 1 + 1/4 + (1*3)/(4*8) + (1*3*5)/(4*8*12) + ... is of the form (1-x)^(-1/2) where x=1/2. Thus, (1 - 1/2)^(-1/2) = (1/2)^(-1/2) = sqrt(2).

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

If  $0 < x , y , a , b < 1 ,$  then the sum of the infinite terms of the series
 $\sqrt { x } ( \sqrt { a } + \sqrt { x } ) + \sqrt { x } ( \sqrt { a b } + \sqrt { x y } ) + \sqrt { x } ( b \sqrt { a } + y \sqrt { x } ) + \ldots$  is

  1. $\dfrac { \sqrt { a x } } { 1 + \sqrt { b } } + \dfrac { x } { 1 + \sqrt { y } }$
  2. $\dfrac { \sqrt { x } } { 1 + \sqrt { b } } + \dfrac { \sqrt { x } } { 1 + \sqrt { y } }$
  3. $\dfrac { \sqrt { x } } { 1 - \sqrt { b } } + \dfrac { \sqrt { x } } { 1 - \sqrt { y } }$
  4. $\dfrac { \sqrt { a x } } { 1 - \sqrt { b } } + \dfrac { x } { 1 - \sqrt { y } }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The given series can be split into two separate infinite geometric series. The first terms involve powers of b and the second involve powers of y, both with common ratios less than 1. Summing each using the infinite geometric series formula a/(1-r) yields the correct expression.

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

If  $S _ { n }$  denotes the sum of the terms in the  $n ^ { t h }$  bracket of the series $( 1 ) + ( 3 + 5 ) + ( 7 + 9 + 11 ) + ( 13 + 15 + 17 + 19 ) + \ldots \ldots , \text { then } \left( S _ { 11 } - S _ { 9 } \right) =$

  1. $362$
  2. $432$
  3. $602$
  4. $632$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The n-th bracket contains n terms, starting from n^2 - n + 1. The sum of the terms in the n-th bracket is n^3. Therefore, S_n = n^3, so S_11 - S_9 = 11^3 - 9^3 = 1331 - 729 = 602.

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

The sum of the infinite terms of the series $\cot^{-1}\left(1^{2}+\dfrac{3}{4}\right)+\cot^{-1}\left(2^{2}+\dfrac{3}{4}\right)+\cot^{-1}\left(3^{2}+\dfrac{3}{4}\right)+..$ is equal to:

  1. $\tan^{-1}\left(1\right)$
  2. $\tan^{-1}\left(2\right)$
  3. $\tan^{-1}\left(3\right)$
  4. $\tan^{-1}\left(4\right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Rewrite the general term inside the inverse cotangent using partial fractions or telescoping properties, noting that cot^-1(x) = tan^-1(1/x). Expressing the denominator as k^2 + 3/4 allows the series to telescope when converted to tangent inverse form.

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

Sum infinite terms of the series $\cot ^ { - 1 } \left( 1 ^ { 2 } + \frac { 3 } { 4 } \right) + \cot ^ { - 1 } \left( 2 ^ { 2 } + \frac { 3 } { 4 } \right) + \cot ^ { - 1 } \left( 3 ^ { 2 } + \frac { 3 } { 4 } \right) + \ldots$ is

  1. $\pi / 4$
  2. $\tan ^ { - 1 } 2$
  3. $\tan ^ { - 1 } 3$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Converting the given series of cot^-1 into tan^-1 terms via tan^-1(1/x) and manipulating the general term enables it to be expressed as a telescoping sum whose limit evaluates to tan^-1(2).

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

The sum of infinity terms of the series $\dfrac{1}{1+1^2+1^4} + \dfrac{1}{1+2^2+2^4} + \dfrac{3}{1+3^2+3^4}+....\infty$ is 

  1. $\dfrac{1}{2}$
  2. $\dfrac{1}{3}$
  3. $1$
  4. $\dfrac{1}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The n-th term is 1 / (1 + n^2 + n^4) = 1 / ((n^2+1)^2 - n^2) = 1 / ((n^2-n+1)(n^2+n+1)). Using partial fractions, this is (1/2) * [ 1/(n^2-n+1) - 1/(n^2+n+1) ]. This is a telescoping series. The sum is 1/2.