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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  $10$  terms of the series  $\left( x + \dfrac { 1 } { x } \right) ^ { 2 } + \left( x ^ { 2 } + \dfrac { 1 } { x ^ { 2 } } \right) ^ { 2 } + \left( x ^ { 3 } + \dfrac { 1 } { x ^ { 3 } } \right) ^ { 2 } + \ldots .$  is

  1. $\left( \dfrac { x ^ { 20 } - 1 } { x ^ { 2 } - 1 } \right) \left( \dfrac { x ^ { 22 } + 1 } { x ^ { 20 } } \right) + 20$
  2. $\left( \dfrac { x ^ { 18 } - 1 } { x ^ { 2 } - 1 } \right) \left( \dfrac { x ^ { 11 } + 1 } { x ^ { 9 } } \right) + 20$
  3. $\left( \dfrac { x ^ { 18 } - 1 } { x ^ { 2 } - 1 } \right) \left( \dfrac { x ^ { 11 } - 1 } { x ^ { 9 } } \right) + 20$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The n-th term is (x^n + 1/x^n)^2 = x^(2n) + 1/x^(2n) + 2. The sum is sum(x^(2n)) + sum(1/x^(2n)) + sum(2). These are geometric series. The sum of x^(2n) is x^2(x^20-1)/(x^2-1).

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

Sum of the series $S=1^{2}-2^{2}+3^{2}-4^{2}+......-2008^{2}+2009^{2}$ is 

  1. $2019045$
  2. $1005004$
  3. $2000506$
  4. $None of these$`
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The series is 1^2 - 2^2 + 3^2 - 4^2 + ... + 2009^2. Grouping terms: (1-2)(1+2) + (3-4)(3+4) + ... + 2009^2 = -(1+2) - (3+4) - ... - (2007+2008) + 2009^2 = -(1+2+3+...+2008) + 2009^2 = -(2008*2009/2) + 2009^2 = 2009 * (2009 - 1004) = 2009 * 1005 = 2019045.

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

the sum of the first n terms of the series ${ 1 }^{ 2 }+{ 2.2 }^{ 2 }+{ 3 }^{ 2 }+{ 2.4 }^{ 2 }+{ 5 }^{ 2 }+{ 2.6 }^{ 2 }....is\frac { n(n+1)^{ 2 } }{ 2 } $ when n is even.wheen n is odd the sum is

  1. $\frac { 3n(n+1) }{ 2 } $
  2. $\frac { { n }^{ 2 }(n+1) }{ 2 } $
  3. $\frac { { n }(n+1)^{ 2 } }{ 4 } $
  4. $\left[ \frac { { n }(n+1) }{ 4 } \right] ^{ 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The series is 1^2 + 2*2^2 + 3^2 + 2*4^2 + 5^2 + 2*6^2 + ... For odd n, the sum follows a specific pattern. By testing n=1, sum=1. By testing n=3, sum=1^2 + 2*2^2 + 3^2 = 1 + 8 + 9 = 18. Formula A: 3*3(4)/2 = 18. This matches.

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

If the sum of first n positive integers is $ \frac{1}{5}  $ times the sum of their squares, then n equals

  1. 5

  2. 6

  3. 7

  4. 8
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The sum of the first n positive integers is n(n+1)/2, and the sum of their squares is n(n+1)(2n+1)/6. Setting n(n+1)/2 = (1/5) * n(n+1)(2n+1)/6 and solving for n yields 2n+1 = 15, so n = 7.

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

Let $\begin{bmatrix} n \ k\end{bmatrix}$ represents the combination of 'n' things taken 'k' at a time, then the value of the sum $\begin{bmatrix} 99\ 97\end{bmatrix}+\begin{bmatrix} 98\ 96\end{bmatrix}+\begin{bmatrix} 97\ 95\end{bmatrix}+...+\begin{bmatrix} 3\ 1\end{bmatrix}+\begin{bmatrix} 2\ 0\end{bmatrix}$ equals?

  1. $\begin{bmatrix} 99\\ 97\end{bmatrix}$
  2. $\begin{bmatrix} 100\\ 98\end{bmatrix}$
  3. $\begin{bmatrix} 99\\ 98\end{bmatrix}$
  4. $\begin{bmatrix} 100\\ 97\end{bmatrix}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a sum of combinations. Using the identity C(n, k) + C(n, k-1) = C(n+1, k), the sum C(99, 97) + C(98, 96) + ... + C(2, 0) can be evaluated. This is a known property of Pascal's triangle diagonals.

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

$\frac { 3 }{ 6 } +\frac { 3.5 }{ 6.9 } +\frac { 3.5.7 }{ 6.9.12 } +...\infty =$

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

The given series can be recognized or derived from a binomial expansion form, specifically related to the expansion of (1 - x)^(-n). Evaluating the specific series yields 3*sqrt(3) - 4.

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

Sum of the series
$P=\dfrac{1}{2\sqrt{1}+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+....+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}$ is

  1. $1/10$
  2. $3/10$
  3. $9/10$
  4. $1/2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Rationalize each term in the series. The general term can be written as 1 / (sqrt(n)(sqrt(n+1))(...)), allowing it to telescope into a simple difference of two terms, giving 9/10.

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

The sum of series $\sec^{-1}\sqrt {2}+\sec^{-1}\dfrac {\sqrt {10}}{3}+\sec^{-1}\dfrac {\sqrt {50}}{7}+...+\sec^{-1}\sqrt {\dfrac {(n^{2}+1)(n^{2}-2n+2)}{(n^{2}-n+1)^{2}}}$ 

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

The terms are sec^-1(sqrt(2)), sec^-1(sqrt(10)/3), etc. This simplifies to tan^-1(1) + tan^-1(1/3) + ... which is a telescoping series of arctangents.