I'll upvote everything
which expression represents "3 more than twice a number"?
3−2n3−2n
2n−32n−3
2n+32n+3
2+n+3

Hey there! Hello!

So, for this problem, I don't really see an answer choice that fits along with this description?? The last one is the closest to the answer I believe it is, but still doesn't fit exactly:

Twice a number = 2n
Three more than = +3
Final equation = 2n+3

Perhaps you have a typo either in your question or your answer problem, or I'm just way too tired. Either way, I hope this helped you out, and feel free to ask me any additional questions you may have. 

–Lamb :-)

The given symbol for the nth triangular number = Tₙ

The given symbol for the nth square number = Qₙ

The given symbol for the nth pentagonal number = Pₙ

The mathematical formula for the nth triangular number, Tₙ = n·(n + 1)/2

The nth square number is given by Qₙ = n²

The nth pentagonal number is given by Pₙ = n·(3·n - 1)/2

(a) T₃.ₙ₋₁ = (3·n - 1)·(3·n - 1 + 1)/2 = (3·n - 1)·(3·n)/2  = 3·(3·n - 1)·(n)/2 = 3·Pₙ

Therefore;

T₍₃.ₙ ₋ ₁₎ = 3·Pₙ

(b) Pₙ - Qₙ = n·(3·n - 1)/2 - n²

Expanding the above equation gives;

n·(3·n - 1)/2 - n² = (3·n² - n - 2·n²)/2 = (n² - n)/2 = n·(n - 1)/2

T₍ₙ ₋ ₁₎ = ((n - 1)· ((n - 1) + 1))/2 =  ((n - 1)· n)/2 = n·(n - 1)/2

∴ Pₙ - Qₙ = n·(3·n - 1)/2 - n² = n·(n - 1)/2 = T₍ₙ ₋ ₁₎

∴ Pₙ - Qₙ = T₍ₙ ₋ ₁₎

Therefore, we get;

P₃.ₙ - Q₃.ₙ = T₍₃.ₙ ₋ ₁₎

Where, T₍₃.ₙ ₋ ₁₎ = 3·Pₙ from (a) above, gives;

P₃.ₙ - Q₃.ₙ = 3·Pₙ

∴ P₃.ₙ - 3·Pₙ = Q₃.ₙ

Plugging in the values, gives;

P₃.ₙ = (3·n)·(3·(3·n) - 1)/2 = (3·n)·((9·n) - 1)/2

3·Pₙ = 3·n·(3·n - 1)/2

P₃.ₙ - 3·Pₙ = (3·n)·((9·n) - 1)/2 - 3·n·(3·n - 1)/2 = (3·n)·(((9·n) - 1) - (3·n - 1))/2

(3·n)·(((9·n) - 1) - (3·n - 1))/2 = (3·n)·(9·n - 1 - 3·n + 1)/2 = (3·n)·(6·n )/2 = 9·n²

Q₃.ₙ = (3·n)² = 9·n²

∴ P₃.ₙ - 3·Pₙ = P₃.ₙ - 3·Pₙ

Step-by-step explanation: