Number Sequences: Problems with Solutions

Solution:
[tex]a_{n+1}-a_n=0[/tex] means that [tex]a_{n+1}=a_n[/tex], therefore [tex]137=a_1=a_2=a_3=...=a_{8999}[/tex]

Solution:
The recurrence relation is [tex]a_{n+2}-2a_{n+1}+a_n=0[/tex] and its characteristic equation is
[tex]r^2-2r+1=0[/tex] with a double root [tex]r_{1,2}=1[/tex]. It means that [tex]a_n=c_1 \times 1^n+c_2 \times n \times 1^n=c_1+n \times c_2[/tex]. But any sequence whose explicit member formula is a linear function is an arithmetic progression. Therefore [tex]a_n[/tex] is the arithmetic progression, defined by [tex]a_1=4[/tex] and [tex]d=a_2-a_1=6-4=2[/tex]. Then [tex]a_{15}=a_1+14 \times d=4+14 \times 2=32[/tex]

Solution:
[tex]a_{19}=\frac{4 \times 19-1}{19}=\frac{75}{19}[/tex]

Solution:
[tex]x_2=\frac{2 \times 2+8}{4+3 \times 2}=\frac{12}{10}=\frac{6}{5}[/tex]

Solution:
[tex]a_5==-3+2^5+5^{5-3}=-3+32+5^2=29+25=54[/tex]

Solution:
[tex]a_{n+1}-a_n=2^{n+1}+n+1-(2^n+n)=2.2^n+n+1-2^n-n=2^n+1>0[/tex], so the sequence is increasing.

Solution:
Substitute [tex]n=1, 2, 3[/tex] in each formula:
[tex]5-3n:\quad 5-3=2, 5-6=-1, 5-9=-4[/tex]
For [tex]5-3n^{2}[/tex] we get [tex]a_{2}=-7[/tex]; for [tex]n^{2}-6n+7[/tex] we get [tex]a_{3}=-2[/tex]; for [tex]2n^{2}-5n+1[/tex] we get [tex]a_{1}=-2[/tex]. Only [tex]a_{n}=5-3n[/tex] fits.

Solution:
[tex]a_{2}=3a_{1}+1=-3+1=-2[/tex]
[tex]a_{3}=3a_{2}+2=-6+2=-4[/tex]
[tex]a_{1}a_{2}a_{3}=(-1)(-2)(-4)=-8[/tex]

Solution:
[tex]2n^{2}-17n=-21\iff 2n^{2}-17n+21=0\iff n=\dfrac{17\pm 11}{4}[/tex]
So [tex]n=7[/tex] or [tex]n=\dfrac{3}{2}[/tex]. The index is a natural number, so [tex]n=7[/tex]. Check: [tex]a_{7}=98-119=-21[/tex].

Solution:
Adding the equalities [tex]a_{2}=a_{1}+2, \ldots, a_{n}=a_{n-1}+n[/tex] gives [tex]a_{n}=1+2+3+\ldots+n=\dfrac{n(n+1)}{2}[/tex].
[tex]a_{35}=\dfrac{35\times 36}{2}=630[/tex]

Solution:
The last copy of the number [tex]k[/tex] stands at position [tex]1+2+\ldots+k=\dfrac{k(k+1)}{2}[/tex].
[tex]k=44:\ 990, k=45:\ 1035, 990\lt 1000\le 1035[/tex]
So the [tex]1000[/tex]-th term is in the block of the number [tex]45[/tex]: [tex]a_{1000}=45[/tex].

Solution:
[tex]a_{1}=S_{1}=9[/tex]. For [tex]n\ge 2[/tex]:
[tex]a_{n}=S_{n}-S_{n-1}=4n^{2}-4(n-1)^{2}=8n-4[/tex]
The formula [tex]8n-4[/tex] gives [tex]4\ne 9[/tex] for [tex]n=1[/tex], so the first term must be written separately (the sequence is not an arithmetic progression).

Solution:
The layers contain [tex]10^{2}, 9^{2}, \ldots, 1^{2}[/tex] balls. Using the formula
[tex]1^{2}+2^{2}+\ldots+n^{2}=\dfrac{n(n+1)(2n+1)}{6}[/tex]
[tex]n=10:\ \dfrac{10\times 11\times 21}{6}=385[/tex]

Solution:
Divide the numerator and the denominator by [tex]7^{n}[/tex]:
[tex]\dfrac{\left(\frac{2}{7}\right)^{n}+1}{\left(\frac{2}{7}\right)^{n}-\frac{1}{7}}\to\dfrac{0+1}{0-\frac{1}{7}}=-7[/tex]

Solution:
By the binomial formula [tex](n+1)^{7}-n^{7}=7n^{6}+21n^{5}+\ldots+1[/tex] – a polynomial of degree [tex]6[/tex].
The numerator [tex](n+1)^{6}=n^{6}+6n^{5}+\ldots+1[/tex] also has degree [tex]6[/tex], so the limit is the ratio of the leading coefficients: [tex]\dfrac{1}{7}[/tex].

Solution:
[tex]1+2+\ldots+n=\dfrac{n(n+1)}{2}[/tex]
[tex]\dfrac{n(n+1)}{2(n+2)}-\dfrac{n}{2}=\dfrac{n^{2}+n-n^{2}-2n}{2(n+2)}=\dfrac{-n}{2(n+2)}=\dfrac{-1}{2\left(1+\frac{2}{n}\right)}\to -\dfrac{1}{2}[/tex]

Solution:
Multiply and divide by the conjugate [tex]\sqrt{5n^{2}+2n}+\sqrt{5n^{2}-2n}[/tex]:
[tex]\dfrac{(5n^{2}+2n)-(5n^{2}-2n)}{\sqrt{5n^{2}+2n}+\sqrt{5n^{2}-2n}}=\dfrac{4}{\sqrt{5+\frac{2}{n}}+\sqrt{5-\frac{2}{n}}}\to\dfrac{4}{2\sqrt{5}}=\dfrac{2\sqrt{5}}{5}[/tex]