Question

1³+3³+5³+....+(2n-1)³=2n²(n+1)²

Answer 1

Step-by-step explanation:

• Given 1³+3³+5³+....+(2n-1)³=2n²(n+1)²
• Adding and subtracting we get
• 2^3 + 4^3 + 6^3 + ………..+ (2n – 1)^3 + (2n)^3 – (2^3 + 4^3 + 6^3+ ---+ (2n)^3 )
• Sum of cube of first n natural number = [n(n + 1)/2]^2
• [2n (2n + 1) / 2]^2 – 2^3 [1^3 + 2^3 + 3^3 + ………+ (2n)^3]
• [n(2n + 1)]^2 – 2^3 [n(n + 1) / 2]^2
• So n^2 (2n + 1)^2 – 2^3 n^2 (n + 1)^2 / 2^2
• So n^2 (2n + 1)^2 – 2n^2 (n + 1)^2
• Taking n^2 as common we get
• So n^2 (2n + 1)^2 - 2(n + 1)^2
• = n^2 ( 4n^2 + 4n + 1 – 2n^2 – 4n – 2)
• = n^2 (2n^2 – 1)

Reference link will be

https://brainly.in/question/3628099

Answer 2

Answer:

Solution:

The nth term of the given series

1³ + 5³ + 7³ +…….. ……….

……………………………………………………………………………………………(1)

is, nth term = (2n - 1)³ = 8n³ - 1 - 3.2n.1 (2n-1) = 8n³ - 12n² +6n - 1

Let the sum up to n terms of the series (1) be denoted by S; then

S = 8 ∑n³ - 12 ∑n² +6 ∑n - ∑1

where

∑n³ = 1³ + 2³ + 3³ +…….. ……….+n³

∑n² = 1² + 2² + 3² +…………….+n²

∑n = 1 + 2 + 3 + 4 +…………….+n

and the summation runs from 1 to n . But we know that

∑n³ = n² (n+1)²/4 ,

∑n² = n(n+1) (2n+1)/6 and

∑n = n(n+1)/2

Substituting the above formulae and noting that ∑1 = 1+1+1+…… n times = n,

S = 8 . n² (n+1)²/4 - 12 . n . (n+1) . (2n+1)/6 + 6 . n(n+1)/2 - n

= 2 n² (n²+2n+1) - 2 (n²+n) . (2n+1) + 3 (n²+n) - n

= 2n^4 + 4n³ + 2n² - 2(2n³ + n² + 2n² + n) + 3n² + 3n - n

= 2n^4 + 4n³ + 2n² - 4n³ - 2n² - 4n² - 2n + 3n² + 3n - n  Cancelling out equal terms,

= 2n^4 - 4n² - 2n + 3n² + 3n - n  Simplifying further

S = 2n²(n+1)² (Proved)