how to solve 2n^2+n-(2(n-1)^2-(n-1))=?
Please help me
Answers
Answer:
a=3 and d=6
using the appropriate method above equation is solved
EXPLANATION.
⇒ aₙ = 2n² + n - [2(n - 1)² - (n - 1)].
As we know that,
Formula of :
⇒ (a - b)² = a² + b² - 2ab.
Using this formula in equation, we get.
Simplify the whole equation, we get.
⇒ aₙ = 2n² + n - [2(n² + 1 - 2n) - n + 1].
⇒ aₙ = 2n² + n - [2n² + 2 - 4n - n + 1].
⇒ aₙ = 2n² + n - [2n² - 5n + 3].
⇒ aₙ = 2n² + n - 2n² + 5n - 3.
⇒ aₙ = 5n + n - 3.
⇒ aₙ = 6n - 3.
As we know that,
Put the value of n = 1 in equation, we get.
⇒ 6(1) - 3.
⇒ 3.
Put the value of n = 2 in equation, we get.
⇒ 6(2) - 3.
⇒ 12 - 3 = 9.
Put the value of n = 3 in equation, we get.
⇒ 6(3) - 3.
⇒ 18 - 3.
⇒ 15.
Put the value of n = 4 in equation, we get.
⇒ 6(4) - 3.
⇒ 24 - 3.
⇒ 21.
⇒ Series = 3, 9, 15, 21 ,,,,,,,,
First term = a = 3.
Common difference = d = b - a = 9 - 3 = 6.
Algebraic expression = 6n - 3.
MORE INFORMATION.
Supposition of terms in an A.P.
(1) = Three terms as : a - d, a, a + d.
(2) = Four terms as : a - 3d, a - d, a + d, a + 3d.
(3) = Five terms as : a - 2d, a - d, a, a + d, a + 2d.