Use mathematical induction to prove the statement is true for all positive integers n. 8 + 16 + 24 + ... + 8n = 4n(n + 1)? Please show work

The sum of the series is Sₙ = n/2 [2·a + (n - 1)·d] where a = 8 and d = 8, therefore 8 + 16 + 24 + ... + 8·n = 4·n·(n + 1)

Step-by-step explanation:

The parameters given are;

8 + 16 + 24 + ... + 8·n = 4·n·(n + 1)

The given series of numbers can be checked to find;

16 - 8 = 24 - 16 = 8

Therefore, the series of numbers is an arithmetic progression with first term = 8, and common difference = 8, we have;

The sum of n terms of an arithmetic progression, Sₙ, is given as follows;

Sₙ = n/2 [2·a + (n - 1)·d]

Where;

a = The first term of the series of numbers = 8

d = The common difference = 8

∴ Sₙ = n/2 × [2×8 + (n - 1)×8] = n [2×8/2 + (n - 1)×8/2]  =  n × [8 + (n - 1)×4]

Sₙ =  n × [8 + (n - 1)×4] =  n × [8 + 4·n - 4] = n × [8  - 4 + 4·n] = n × [4 + 4·n]

Sₙ =n × [4 + 4·n] = 4 × n×(n + 1) = 4·n·(n + 1).

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step-by-step explanation:

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb