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}

Substituting \{a_n\} to the given equation

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


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

Simplifying the equation yields


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


This is the typical form of recursive sequence


that could be converted to the following form


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

c= \dfrac{q}{p-1}

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


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}


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

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


Simplifying the expression gives


Steven Zheng posted 1 week ago

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