How to deduce the general term formula of Fibonacci sequence? An = {[(1 + √5)/2]^n - [(1 - √5)/2]^n}/√5

How to deduce the general term formula of Fibonacci sequence? An = {[(1 + √5)/2]^n - [(1 - √5)/2]^n}/√5

From an + 2 = an + 1 + an
There is an + 2-An + 1-an = 0
The characteristic equation x2-x-1 = 0 is constructed,
Let its two roots be p, Q have PQ = - 1, P + q = 1
Let's prove that {an + 1-pan} is an equal ratio sequence with Q as the common ratio
For the convenience of derivation, let A0 = 1, still satisfy an + 2 = an + 1 + an
an+1-pan
= an+an-1 -pan
= (1-p) an-pqan-1
=q(an-pan-1)
So: {an + 1-pan} is an equal ratio sequence with Q as the common ratio
a1-pa0
=1-p=q
So an + 1-pan = Q * QN = QN + 1
Similarly, an + 1-qan = P * PN = PN + 1
①-②:(q-p)an= qn+1-pn
Because P = (1 - √ 5) / 2, q = (1 + √ 5) / 2, Q-P = √ 5, so
an=(1/√5){[(1+√5)/2]n+1-[(1-√5)/2] n+1}
It can be verified that A0 and A1 are also suitable for the above general formula

Finding the general term formula of Fibonacci sequence

Its general formula is: (1 / √ 5) * {[(1 + √ 5) / 2] ^ n - [(1 - √ 5) / 2] ^ n} [√ 5 denotes radical 5]

General term formula of Fibonacci sequence,

The inventor of Fibonacci series is Italian mathematician Leonardo Fibonacci, who was born in 1170 AD and died in 1240 ad. his native place is probably Pisa. He is known as "Leonardo of Pisa". In 1202, he wrote "the principle of abacus"