find the sum of the n terms of the series whose nth term is
4n (n^2+1)-(6n^2+1)
4n (n^2+1)-(6n^2+1)
-
i) Given nth term is: 4n³ - 6n² + 4n - 1
Hence sum to n terms of this series is: S = ∑(4n³ - 6n² + 4n - 1) for n = 1 to n.
[What do you mean be 'harmonical progression'? My knowledge is limited to 'harmonic progression' only; however the given one is not harmonic progression too; it is some defined sequence, whose sum to n terms existing and can be computed in a methodical way; if it is to be a harmonic progression, kindly note its sum to n terms cannot be found with any defined formulas].
ii) Coming back to the question, taking its sum as given above:
S = ∑(4n³ - 6n² + 4n - 1) for n = 1 to n.
Splitting each of them, S = 4∑n³ - 6∑n² + 4∑n - ∑1 [all for n = 1 to n]
4∑n³ = 4(1³ + 2³ + 3³ + 4³ + ........ + n³) = 4{n(n + 1)/2}² = 4(n²)(n+1)²/4 = n²(n + 1)²
[Sum of cubes of 1st n natural numbers = {n(n + 1)/2}²]
-6∑n² = -6(1² + 2² + 3² + ....... + n²) = -6n(n + 1)(2n + 1)/6 = -n(n + 1)(2n + 1)
[Sum of squares of 1st n natural numbers = n(n + 1)(2n + 1)/6]
4∑n = 4( 1 + 2 + 3 + ..... + n) = 4n(n + 1)/2 = 2n( n + 1)
- ∑1 = -(1 + 1 + 1 + ...) = -n
Adding all the above,
S = n²(n + 1)² - n(n + 1)(2n + 1) + 2n(n + 1) - n
Simplifying this, S = n^4.
Check:
i) For n = 1, S = 1; also 1st term is 4*1(1 + 1) - (6 + 1) = 1
ii) For n = 2, S = 2^4 = 16; 1st + 2nd = 1 + 15 = 16
iii) For n = 3, S = 3^4 = 81; 1st + 2nd + 3rd = 1 + 15 + 65 = 81
Hence sum to n terms of this series is: S = ∑(4n³ - 6n² + 4n - 1) for n = 1 to n.
[What do you mean be 'harmonical progression'? My knowledge is limited to 'harmonic progression' only; however the given one is not harmonic progression too; it is some defined sequence, whose sum to n terms existing and can be computed in a methodical way; if it is to be a harmonic progression, kindly note its sum to n terms cannot be found with any defined formulas].
ii) Coming back to the question, taking its sum as given above:
S = ∑(4n³ - 6n² + 4n - 1) for n = 1 to n.
Splitting each of them, S = 4∑n³ - 6∑n² + 4∑n - ∑1 [all for n = 1 to n]
4∑n³ = 4(1³ + 2³ + 3³ + 4³ + ........ + n³) = 4{n(n + 1)/2}² = 4(n²)(n+1)²/4 = n²(n + 1)²
[Sum of cubes of 1st n natural numbers = {n(n + 1)/2}²]
-6∑n² = -6(1² + 2² + 3² + ....... + n²) = -6n(n + 1)(2n + 1)/6 = -n(n + 1)(2n + 1)
[Sum of squares of 1st n natural numbers = n(n + 1)(2n + 1)/6]
4∑n = 4( 1 + 2 + 3 + ..... + n) = 4n(n + 1)/2 = 2n( n + 1)
- ∑1 = -(1 + 1 + 1 + ...) = -n
Adding all the above,
S = n²(n + 1)² - n(n + 1)(2n + 1) + 2n(n + 1) - n
Simplifying this, S = n^4.
Check:
i) For n = 1, S = 1; also 1st term is 4*1(1 + 1) - (6 + 1) = 1
ii) For n = 2, S = 2^4 = 16; 1st + 2nd = 1 + 15 = 16
iii) For n = 3, S = 3^4 = 81; 1st + 2nd + 3rd = 1 + 15 + 65 = 81