#### Question

If the sequence \{a_n\} has its first term a_1=1, a_{n+1}=\dfrac{1+4a_n+\sqrt{1+24a_n}}{16} , find the formula for general terms of the sequence.

Collected in the board: Recursive Sequence

Steven Zheng posted 1 week ago

Let b_n=\sqrt{1+24a_n},

Then b_1=\sqrt{1+24\cdotp a_1 }= 5

Since square root of any real numbers are larger than or equal to zero, b_n\geq 0 Taking the square of the equation to express a_n in terms of b_n, we get

a_n= \dfrac{b^2_n-1}{24}
(1)

Substituting \{a_n\} to the given equation

a_{n+1}= \dfrac{1+4a_n+\sqrt{1+24a_n}} {16}

gives

\dfrac{b^2_{n+1}-1}{24}=\dfrac{1+4\cdotp \dfrac{b^2_n-1}{24} +b_n}{16}

Simplifying the equation yields

(2b_{n+1})^2=(b_n+3)^2

Since \{b_n\}\geq 0 , taking the square root of the above equation would not change the sign of two sides. That is,

b_{n+1}=\dfrac{1}{2}b_n+\dfrac{3}{2}
(2)

This is the typical form of recursive sequence

z_{n+1}=pz_n+q

that could be converted to the following form

z_{n+1}+c=p(z_n+c)

In which, c is a constant dependent on the values of p and q

c= \dfrac{q}{p-1}
(3)

In the expression (2),

p=\dfrac{1}{2} and q=\dfrac{3}{2}

Plug in the expression (3)

c=\dfrac{\dfrac{3}{2} }{\dfrac{1}{2}-1 } =-3

Therefore, tte equation (2) is transformed to

b_{n+1}-3=\dfrac{1}{2}(b_n-3)

in which b_n-3 is a geometric sequence with its first term as b_1-3=2, common ratio \dfrac{1}{2} and the formula of general terms

b_n-3=2\cdotp (\dfrac{1}{2})^{n-1}

Therefore,

b_n= 3+(\dfrac{1}{2})^{n-2}
(4)

Substituting b_n to (1) yields the formula of general terms for a_n.

a_n=\dfrac{(3+(\dfrac{1}{2})^{n-2})^2-1}{24}

Simplifying the expression gives

a_n=\dfrac{2}{3}(\dfrac{1}{2})^{2n}+(\dfrac{1}{2})^n+\dfrac{1}{3}

Steven Zheng posted 1 week ago

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