find the formula for the general term aₙ of the sequence starting at a1

Question

find the formula for the general term aₙ of the sequence starting at a1

{-4/25, 5/36, -6/49, 7/64, ...}

aₙ =

Answer 1

it alternates sign, so you need a (-1)^n term in the general formula. a1 is negative a3 is negative, a2, a4, and even terms are positive

the general term is a fraction where the numerator increases by 1 each term

first term is 4, then add 1 each term

the denominator is a perfect square of 1 + the numerator a1 is -4/(4+1)^2

a2 is 5(5+1)^2, a3 is -6/(6+1)^2

an = (-1)^n[(4+n-1)/(4+n)^2 = (-1)^n[(n+3)/(n+4)^2]

a1 = (-1)^1[(1+3)/(1+4)^2] = -4/25

a2 = (-1)^2[(2+3)/(2+4)^2] = 5/36

an = (-1)^n[(n+3)/(n+4)^2