Tag: some special sequences

Questions Related to some special sequences

The expression $(x + 1)(x + 2)(x + 3)(x + 4) + 1$ is a 

maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

  1. perfect square

  2. cube

  3. quartic polynomial

  4. none of the above

Show answer
Correct Option: A
Explanation:

Solving

$(x+1)(x+2)(x+3)(x+4)+1$
$Multiplying\ first\ bracket\ with\ last\ and\ second\ to\ the\ third\ one$
$(x^2+5x+4)(x^2+5x+6)+1$
$Replacing\ x^2+5x+4\ by\ 'B'$
$(B)(B+2)+1$
$B^2+2B+1$
$(B+1)^2=(x^2+5x+5)^2$
Hence $L.H.S.$ is the $Perfect\ Square$ of $(x^2+5x+5)$



$\cfrac { { \left( 963+476 \right)  }^{ 2 }+{ \left( 963-476 \right)  }^{ 2 } }{ \left( 973\times 963+476\times 476 \right)  } =$?

maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

  1. $1449$

  2. $497$

  3. $2$

  4. $4$

  5. None of these

Show answer
Correct Option: C
Explanation:

Given Exp.$=\cfrac { { \left( a+b \right)  }^{ 2 }+{ \left( a-b \right)  }^{ 2 } }{ \left( { a }^{ 2 }+{ b }^{ 2 } \right)  } =\cfrac { 2\left( { a }^{ 2 }+{ b }^{ 2 } \right)  }{ \left( { a }^{ 2 }+{ b }^{ 2 } \right)  } =2$

By what least number $21600$ must be multiplied to make it a perfect cube?

maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

  1. $6$

  2. $10$

  3. $30$

  4. $60$

Show answer
Correct Option: B
Explanation:

$21600 $ can be factorized as $6\times 6\times 6\times 10\times 10$

To make it perfect cube, it must be multiplied by $10$.

A square is inscribed in the circle $x^2+y^2-10x- 6y +30=0$. One side of the square is parallel to $y=x+3$. Then which of the following can be a vertex of the square

maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

  1. $(3, 3)$

  2. $(7, 3)$

  3. $(5, 5)$

  4. $(1, 1)$

Show answer
Correct Option: A,B

For real number $a,b,c$ and $d$ , if $a^2+b^2=4$ and $c^2+d^2=1$, then possible value of $ac+bd$ is / are 

maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

  1. $2$

  2. $3$

  3. $1$

  4. $\dfrac{1}{4}$

Show answer
Correct Option: A,C,D
Explanation:

According to Cauchy-Schwarz inequality:


$(ac+bd)^2\le(a^2+b^2)(c^2+d^2)$      $(\forall a, b, c, d \in \mathbb{R})$

Substituting the values here gives:

$(ac+bd)^2\le4$

$(ac+bd)^\le2$

Therefore, it can take all values given in the options except 3.

What is the least number that must be added to $594$ to make sum a perfect square?

maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

  1. $13$

  2. $29$

  3. $31$

  4. $33$

Show answer
Correct Option: C
Explanation:

First calculate the square-root of $594$

$\sqrt{594}\approx 24.37$

The whole number larger than $24.37$ is $25$

and $(25)^{2}=625$

Now, $625$ is a perfect square.

So, the least number that must be added to $594$ to make sum a perfect square is $=625-594=31$

A rectangle with integer side length has perimeter $10$. What is the greatest numbers of these rectangles that can be cut from a piece of paper with width $24$ and length $60$?

maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

  1. $144$

  2. $180$

  3. $240$

  4. $360$

  5. $480$

Show answer
Correct Option: D
Explanation:

If a rectangle has perimeter 10 then the sum of its length and width is 5, giving two choice with integer sides:

$(i)2\times3$ rectangle of area $6$
$(ii)1\times4$ rectangle of area $4$
The piece of paper has area $24\times60=1400$
This can be divided into $12\times20=240$ rectangles with sides $2\times3$
It can be divided into $24\times15=360$ recatngles with sides $1\times4$
So, the greatest number of rectangles is $360$

Fourth roots of $193-4\sqrt{2178}$ is

maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

  1. $(7-\sqrt{2})$

  2. $(5-\sqrt{2})$

  3. $(3-\sqrt{2})$

  4. $(10-\sqrt{7})$

Show answer
Correct Option: C
Explanation:
According to Question

$(193-4\sqrt{2178} )^{1/4}$

$=(193-4\sqrt{11\times11\times3\times3\times2} )^{1/4}$

$=(193-4\times11\times3\sqrt2 )^{1/4}$

$=(121+72-132\sqrt{2} )^{1/4}$

$=(11^2+(6\sqrt2)^2-2\times11\times6\sqrt2)^{1/4}$                          $Using\ a^2+b^2-2ab=(a-b)^2$

$=(11-6\sqrt2)^{2\times0.25}$

$=(9+2-6\sqrt{2})^{0.5}$

$=(3^2+\sqrt{2}\ ^2-2\times3\times\sqrt{2})^{0.5}$                          $Using\ a^2+b^2-2ab=(a-b)^2$

$=(3-\sqrt{2})^{0.5\times2}$

$=3-\sqrt{2}$

$C$ is the right answer

The value of $1^{2}+3^{2}+5^{2}+.....25^{2}$ is:

maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

  1. $1728$

  2. $1456$

  3. $2925$

  4. $1469$

Show answer
Correct Option: C
Explanation:

$1^2+3^2+5^2+....+25^2$

$\Rightarrow$  $(1^2+2^2+3^2+4^2+....+25^2)-(2^2+4^2+6^2+8^2....24^2)$

$\Rightarrow$  $(1^2+2^2+3^2+4^2+...25^2)-[(2\times 1)^2+(2\times 2)^2+(2\times 3)^2+....(2\times 12)^2]$

$\Rightarrow$  $(1^2+2^2+3^2+....+25^2)-2^2(1^2+2^2+3^2+.....12^2)$

$1^2+2^2+3^2+...+n^2=\dfrac{n(n+1)(2n+1)}{6}$

$\Rightarrow$  $\dfrac{25(25+1)(2\times 25+1)}{6}-2^2\dfrac{12(12+1)(2\times 12+1)}{6}$

$\Rightarrow$  $\dfrac{25\times 26\times 51}{6}-4\times\dfrac{12\times 13\times 25}{6}$

$\Rightarrow$  $25\times 13\times 17-4\times 2\times 13\times 25$

$\Rightarrow$  $25\times 13(17-8)$

$\Rightarrow$  $25\times 13\times 9$

$\Rightarrow$  $2925$

Is it possible for the square of a number to end with 5 zeroes?
State true or false.

maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

The square of a number with 'n' number of zeroes, will have $ 2 \times n $ zeroes. Hence, there cannot be $ 5 $ zeroes in a square of a number as it would be an even number.