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kumpel [21]
2 years ago
12

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Mathematics
1 answer:
Bezzdna [24]2 years ago
3 0

Since a_1,a_2,a_3,\cdots are in arithmetic progression,

a_2 = a_1 + 2

a_3 = a_2 + 2 = a_1 + 2\cdot2

a_4 = a_3+2 = a_1+3\cdot2

\cdots \implies a_n = a_1 + 2(n-1)

and since b_1,b_2,b_3,\cdots are in geometric progression,

b_2 = 2b_1

b_3=2b_2 = 2^2 b_1

b_4=2b_3=2^3b_1

\cdots\implies b_n=2^{n-1}b_1

Recall that

\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n

\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2

It follows that

a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n +  n(n-1)

so the left side is

2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n

Also recall that

\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}

so that the right side is

b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)

Solve for c.

2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}

Now, the numerator increases more slowly than the denominator, since

\dfrac{d}{dn}(2n(n-1)) = 4n - 2

\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2

and for n\ge5,

2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2

This means we only need to check if the claim is true for any n\in\{1,2,3,4\}.

n=1 doesn't work, since that makes c=0.

If n=2, then

c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0

If n=3, then

c = \dfrac{12}{2^3 - 6 - 1} = 12

If n=4, then

c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N

There is only one value for which the claim is true, c=12.

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