Hire us
Support us
(New)
All problem can be solved using search box
I want to sell my website www.AtoZmath.com with complete code
Home
College Algebra
Games
Feedback
Support us
About us
Algebra
Matrix Algebra
Numerical Methods
Statistical Methods
Operation Research
Word Problems
Calculus
Geometry
Pre-Algebra
Translate this page
What's new
1.
All problem solution using search box on 05.10.18
2.
Share solution on facebook, twitter with your friends on 17.08.18
3.
Matrix Structure
on 24.07.18
4.
Algorithm, Formulas, Examples added at respective pages on 20.07.18
Topics
Home
Algebra
Matrix Algebra
Numerical Methods
Statistical Methods
Operation Research
Word Problems
Calculus
Geometry
Pre-Algebra
College Algebra
Games
Test
Method and examples
Geometric Progression
1. For given geometric progression series 3,6,12,24,48 ,... find 10 th term and addition of first 10 th terms.
2. For given geometric progression series 3,6,12,24,48 ,... then find n such that S(n) = 3069 .
3. For given geometric progression series 3,6,12,24,48 ,... then find n such that f(n) = 1536 .
4. For geometric progression f( 1 ) = 2 , f( 4 ) = 54 then find f( 3 ) and S( 3 ).
5. For geometric progression f( 1 ) = 2 , f( 4 ) = 54 , then find n such that f(n) = 18 .
6. For geometric progression addition of 3 terms is 26 and their multiplication is 216 , then that numbers
7. For geometric progression multiplication of 5 terms is 1 and 5 th term is 81 times then the 1 th term.
8. Arithmetic mean of two number is 13 and geometric mean is 12 , then find that numbers
9. Two numbers are in the ratio 9 : 16 and difference of arithmetic mean and geometric mean is 1 , then find that numbers
10. Find 6 arithmetic mean between 3 and 24 .
11. Find 3 geometric mean between 1 and 256 .
12. Prove that 1 + (1 + 2) + (1 + 2 + 3) + ... n terms = n/6 (n + 1) (n + 2)
13. Prove that 1 * (2^2 - 3^2) + 2 * (3^2 - 4^2) + 3 * (4^2 - 5^2) + ... n terms = n/6 (n + 1) (4n + 11)
14. 1 + x^4 + 3^2 + 4 + x^6 + 6^2 + 7 + x^8 + 9^2 + ... 3n terms
15. 1 + (1 + 3) + (1 + 3 + 5) + ... n terms
16. Prove that for all n belongs to N, 1^2 * n + 2^2 * (n - 1) + 3^2 * (n - 2) + ... + n^2 * 1 = n/12 (n + 1)^2 (n + 2)
17. Prove that 1 * 2^2 + 3 * 5^2 + 5 * 8^2 + ... n terms = n/2 (9n^3 + 4n^2 - 4n - 1)
18. Prove that 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ... n terms = n/12 (n + 1)^2 (n + 2)
19. Prove that 2 + 5 + 10 + 17 + ... n terms = n/6 (2n^2 + 3n + 7)
20. Prove that sum [ sum (2n -3) ] = n/6 (n + 1)(2n - 5)
21. For geometric progression, find 1 + 1/sqrt(2) + 1/2 + 1/(2*sqrt(2)) + ... 10 terms ( For geometric progression, find 1 + ...
22. Find 1^2 + 2^2 + ...+ 10^2 , ( Find a^2 + b^2 + ... + n^2 , where a = 1 , b = 2 and n = 10 . )
23. Find 1^2 + 2^2 + ... 10 terms, ( Find a^2 + b^2 + ... n terms, where a = 1 , b = 2 and n = 10 . )
Problem 13 of 23
13. Prove that `1 * (2^2 - 3^2) + 2 * (3^2 - 4^2) + 3 * (4^2 - 5^2) + ... n` terms `= n/6 (n + 1) (4n + 11)`
Solution
Example
Solution
Solution provided by AtoZmath.com
Want to know about
AtoZmath.com and me
This is demo example. Please click on Find button and solution will be displayed in Solution tab (step by step)
Geometric Progression
13. Prove that `1 * (2^2 - 3^2) + 2 * (3^2 - 4^2) + 3 * (4^2 - 5^2) + ... n` terms `= n/6 (n + 1) (4n + 11)`
L.H.S. `= 1 × (2^2 - 3^2) + 2 × (3^2 - 4^2) + 3 × (4^2 - 5^2) + ... n` terms
`= sum [ f(n) ]`
`= sum [ n ((n + 1)^2 - (n + 2)^2) ]`
`= sum [ n (n^2 + 2n + 1 - n^2 - 2n - 4) ]`
`= sum [ n (-2n - 3) ]`
`= sum (-2n^2 - 3n) ]`
`= -2 sum n^2 - 3 sum n`
`= -2 * (n (n + 1) (2n + 1))/6 - 3 * (n (n + 1))/2`
`= - n/6 (n + 1) [ 2(2n + 1) + 9 ]`
`= - n/6 (n + 1) (4n + 11)`
`=` R.H.S. (Proved)
Share with your friends
Home
College Algebra
Games
Feedback
Support us
About us
Copyright © 2018. All rights reserved.
Terms
,
Privacy
Review Consent