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Toolkit 73

3 Approaches for Finding the Closed Form of aₙ

73.1

an=an−1+3n−4,n≥1a_n=a_{n-1}+3n-4,\qquad n\ge 1
a0=7a_0=7

Write the consecutive equations:

an−an−1=3n−4a_n-a_{n-1}=3n-4
an−1−an−2=3n−7a_{n-1}-a_{n-2}=3n-7
an−2−an−3=3n−10a_{n-2}-a_{n-3}=3n-10
⋮\vdots
a2−a1=3(2)−4a_2-a_1=3(2)-4
a1−a0=3(1)−4a_1-a_0=3(1)-4

Add these equations so that the intermediate terms telescope:

an−a0=((3(1)−4)+(3n−4))n2=(3n−5)n2.a_n-a_0=\frac{((3(1)-4)+(3n-4))n}{2}=\frac{(3n-5)n}{2}.

Therefore,

an=n(3n−5)2+a0=n(3n−5)2+7.a_n=\frac{n(3n-5)}{2}+a_0=\boxed{\frac{n(3n-5)}{2}+7}.

73.2

If the coefficient of an−1a_{n-1} is a constant different from 11:

an=2an−1+5,n≥1a_n=2a_{n-1}+5,\qquad n\ge 1
a0=9a_0=9

Let

bn=an2n.b_n=\frac{a_n}{2^n}.

From

an=2an−1+5,a_n=2a_{n-1}+5,

divide by 2n2^n:

an2n=an−12n−1+52n.\frac{a_n}{2^n}=\frac{a_{n-1}}{2^{n-1}}+\frac{5}{2^n}.

Therefore,

bn=bn−1+52n,n≥1,b_n=b_{n-1}+\frac{5}{2^n},\qquad n\ge 1,

and

b0=a020=9.b_0=\frac{a_0}{2^0}=9.

Now write:

bn−bn−1=52nb_n-b_{n-1}=\frac{5}{2^n}
bn−1−bn−2=52n−1b_{n-1}-b_{n-2}=\frac{5}{2^{n-1}}
bn−2−bn−3=52n−2b_{n-2}-b_{n-3}=\frac{5}{2^{n-2}}
⋮\vdots
b2−b1=522b_2-b_1=\frac{5}{2^2}
b1−b0=521.b_1-b_0=\frac{5}{2^1}.

Adding gives

bn−b0=521+522+⋯+52n−1+52n.b_n-b_0=\frac{5}{2^1}+\frac{5}{2^2}+\cdots+\frac{5}{2^{n-1}}+\frac{5}{2^n}.

Then

bn−b0=52(1+12+⋯+12n−2+12n−1).b_n-b_0=\frac{5}{2}\left(1+\frac{1}{2}+\cdots+\frac{1}{2^{n-2}}+\frac{1}{2^{n-1}}\right).

Using the geometric series,

bn−b0=52⋅1−12n1−12=5(1−12n).b_n-b_0=\frac{5}{2}\cdot\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}=5\left(1-\frac{1}{2^n}\right).

Thus,

bn=5(1−12n)+b0=14−52n.b_n=5\left(1-\frac{1}{2^n}\right)+b_0=14-\frac{5}{2^n}.

Since

an=2nbn,a_n=2^n b_n,

we get

an=14⋅2n−5.\boxed{a_n=14\cdot 2^n-5}.

73.3

If the coefficient of an−1a_{n-1} is nn:

an=nan−1+(n−2)!,n≥2a_n=na_{n-1}+(n-2)!,\qquad n\ge 2
a1=5a_1=5

Let

bn=ann!.b_n=\frac{a_n}{n!}.

From

an=nan−1+(n−2)!,a_n=na_{n-1}+(n-2)!,

divide by n!n!:

ann!=an−1(n−1)!+(n−2)!n!.\frac{a_n}{n!}=\frac{a_{n-1}}{(n-1)!}+\frac{(n-2)!}{n!}.

Therefore,

bn=bn−1+1n(n−1)=bn−1+1n−1−1n,n≥2.b_n=b_{n-1}+\frac{1}{n(n-1)}=b_{n-1}+\frac{1}{n-1}-\frac{1}{n},\qquad n\ge 2.

Also,

b1=a11!=5.b_1=\frac{a_1}{1!}=5.

Now write:

bn−bn−1=1n−1−1nb_n-b_{n-1}=\frac{1}{n-1}-\frac{1}{n}
bn−1−bn−2=1n−2−1n−1b_{n-1}-b_{n-2}=\frac{1}{n-2}-\frac{1}{n-1}
bn−2−bn−3=1n−3−1n−2b_{n-2}-b_{n-3}=\frac{1}{n-3}-\frac{1}{n-2}
⋮\vdots
b3−b2=12−13b_3-b_2=\frac{1}{2}-\frac{1}{3}
b2−b1=1−12.b_2-b_1=1-\frac{1}{2}.

After telescoping,

bn−b1=1−1n.b_n-b_1=1-\frac{1}{n}.

Therefore,

bn=1−1n+b1=6−1n.b_n=1-\frac{1}{n}+b_1=6-\frac{1}{n}.

Hence,

an=n! bn=6n!−(n−1)!.a_n=n!\,b_n=6n!-(n-1)!.

So,

an=6n!−(n−1)!.\boxed{a_n=6n!-(n-1)!}.
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