Home
Class 12
MATHS
If the sum of the first 15 terms of the ...

If the sum of the first 15 terms of the series `((3)/(4))^(3)+(1(1)/(2))^(3)+(2(1)/(4))^(3)+3^(3)+(3(3)/(4))^(3)+...` is equal to 225k, then k is equal to

A

9

B

27

C

108

D

54

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the value of \( k \) given that the sum of the first 15 terms of the series \[ \left(\frac{3}{4}\right)^3 + \left(\frac{3}{2}\right)^3 + \left(\frac{9}{4}\right)^3 + 3^3 + \left(\frac{15}{4}\right)^3 + \ldots \] is equal to \( 225k \). ### Step 1: Identify the pattern in the series The terms of the series can be rewritten as: - \( \left(\frac{3}{4}\right)^3 \) - \( \left(\frac{3}{2}\right)^3 = \left(\frac{6}{4}\right)^3 \) - \( \left(\frac{9}{4}\right)^3 = \left(\frac{9}{4}\right)^3 \) - \( 3^3 = \left(\frac{12}{4}\right)^3 \) - \( \left(\frac{15}{4}\right)^3 \) We can see that the series can be expressed in terms of \( \left(\frac{3n}{4}\right)^3 \) for \( n = 1, 2, 3, \ldots, 15 \). ### Step 2: Rewrite the series Thus, we can express the sum of the first 15 terms as: \[ \sum_{n=1}^{15} \left(\frac{3n}{4}\right)^3 \] ### Step 3: Factor out \( \left(\frac{3}{4}\right)^3 \) Factoring out \( \left(\frac{3}{4}\right)^3 \): \[ \left(\frac{3}{4}\right)^3 \sum_{n=1}^{15} n^3 \] ### Step 4: Use the formula for the sum of cubes The formula for the sum of the first \( n \) cubes is: \[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \] For \( n = 15 \): \[ \sum_{n=1}^{15} n^3 = \left(\frac{15 \cdot 16}{2}\right)^2 = (120)^2 = 14400 \] ### Step 5: Substitute back into the equation Now substituting back into our expression: \[ \left(\frac{3}{4}\right)^3 \cdot 14400 \] Calculating \( \left(\frac{3}{4}\right)^3 \): \[ \left(\frac{3}{4}\right)^3 = \frac{27}{64} \] Thus, the sum becomes: \[ \frac{27}{64} \cdot 14400 \] ### Step 6: Simplify the expression Calculating \( \frac{27 \cdot 14400}{64} \): First, simplify \( 14400 \div 64 \): \[ 14400 \div 64 = 225 \] Now, multiply by 27: \[ 27 \cdot 225 = 6075 \] ### Step 7: Set the equation equal to \( 225k \) We have: \[ 6075 = 225k \] ### Step 8: Solve for \( k \) Dividing both sides by 225: \[ k = \frac{6075}{225} = 27 \] Thus, the value of \( k \) is \( \boxed{27} \). ---
Promotional Banner

Topper's Solved these Questions

  • JEE 2019

    CENGAGE ENGLISH|Exercise Chapter 1|5 Videos
  • JEE 2019

    CENGAGE ENGLISH|Exercise Chapter 2|7 Videos
  • JEE 2019

    CENGAGE ENGLISH|Exercise chapter -3|1 Videos
  • INVERSE TRIGONOMETRIC FUNCTIONS

    CENGAGE ENGLISH|Exercise Archives (Numerical Value type)|2 Videos
  • LIMITS

    CENGAGE ENGLISH|Exercise Comprehension Type|4 Videos

Similar Questions

Explore conceptually related problems

The sum of the first 10 terms of the series (5)/(1.2.3)+(7)/(2.3.9)+(9)/(3.4.27)+….. is

The sum of first 9 terms of the series (1^(3))/(1)+(1^(3)+2^(3))/(1+3)+(1^(3)+2^(3)+3^(3))/(1+3+5)+"........" is

Sum to n terms of the series 1^(3) - (1.5)^(3) +2^(3)-(2.5)^(3) +…. is

The sum of the series 1 + (1)/(1!) ((1)/(4)) + (1.3)/(2!) ((1)/(4))^(2) + (1.3.5)/(3!) ((1)/(4))^(3)+ ... to infty , is

The sum of 10 terms of the series 1+2(1.1)+3(1.1)^(2)+4(1.1)^(3)+…. is

If the sum of the first ten terms of the series (1 3/5)^2+(2 2/5)^2+(3 1/5)^2+4^2+(4 4/5)^2+. . . . . , is (16)/5 m, then m is equal to: (1) 102 (2) 101 (3) 100 (4) 99

The sum of the first 9 terms of the series 1^3/1 + (1^3 + 2^3)/(1+3) + (1^3 + 2^3 +3^3)/(1 + 3 +5) ..... is :

FInd the sum of infinite terms of the series 1/(1.2.3)+1/(2.3.4)+1/(3.4.5).....

If the surm of the first ten terms of the series, (1 3/5)^2+(2 2/5)^2+(3 1/5)^2+4^2+(4 4/5)^2+........ , is 16/5m ,then m is equal to

Find the sum to n terms of the series (1.2.3) + (2.3.4) + (3.4.5) ...

CENGAGE ENGLISH-JEE 2019-MCQ
  1. The product of three consecutive terms of a GP is 512. If 4 is added t...

    Text Solution

    |

  2. Let S(k) = (1+2+3+...+k)/(k). If S(1)^(2) + s(2)^(2) +...+S(10)^(2) = ...

    Text Solution

    |

  3. If the sum of the first 15 terms of the series ((3)/(4))^(3)+(1(1)/(2)...

    Text Solution

    |

  4. Let x, y be positive real numbers and m, n be positive integers, The ...

    Text Solution

    |

  5. Consider a class of 5 girls and 7 boys. The number of different teams ...

    Text Solution

    |

  6. Let S be the set of all triangles in the xy-plane, each having one ver...

    Text Solution

    |

  7. The number of natural numbers less than 7,000 which can be formed by u...

    Text Solution

    |

  8. If set A={1, 2, 3, ?2, }, then the find the number of onto functions f...

    Text Solution

    |

  9. Let S={1,2,3, …, 100}. The number of non-empty subsets A to S such tha...

    Text Solution

    |

  10. Consider three boxes, each containing 10 balls labelled 1, 2, …,10. Su...

    Text Solution

    |

  11. Let Z be the set of integers. If A = {x in Z : 2^((x + 2)(x^(2) - 5x +...

    Text Solution

    |

  12. There are m men and two women participating in a chess tournament. Eac...

    Text Solution

    |

  13. If the fractional part of the number (2^(403))/(15) is (k)/(15) then k...

    Text Solution

    |

  14. The coefficient of t^4 in ((1-t^6)/(1-t))^3 (a) 18 (b) 12 ...

    Text Solution

    |

  15. If Sigma(i=1)^(20) ((""^(20)C(i-1))/(""^(20)C(i)+""^(20)C(i-1)))^(3)=(...

    Text Solution

    |

  16. If the third term in expansion of (1+x^(log2x))^5 is 2560 then x is eq...

    Text Solution

    |

  17. The positive value of lambda for which the coefficient of x^(2) in the...

    Text Solution

    |

  18. If Sigma(r=0)^(25) (""^(50)C(r)""^(50-r)C(25-r))=K(""^(50)C(25)), th...

    Text Solution

    |

  19. If the middle term of the expansion of (x^3/3+3/x)^8 is 5670 then sum ...

    Text Solution

    |

  20. The value of r for which .^(20)C(r ), .^(20)C(r - 1) .^(20)C(1) + .^...

    Text Solution

    |