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$ (1.)\:\: AS_n = a + d(n - 1) \\[5ex] (2.)\:\: AS_n = vn + w \:\:where\:\: v = d \:\:and\:\: w = a - d \\[5ex] (3.)\:\: p = a + d(n - 1) \\[5ex] (4.)\:\: SAS_n = \dfrac{n}{2}(a + AS_n) \\[7ex] (5.)\:\: SAS_n = \dfrac{n}{2}(a + p) \\[7ex] (6.)\:\: SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[7ex] (7.)\:\: n = \dfrac{2 * SAS_n}{a + p} \\[7ex] (8.)\:\: n = \dfrac{p - a + d}{d} \\[7ex] (9.)\:\: n = \dfrac{-(2a - d) \pm \sqrt{(2a - d)^2 + 8d*SAS_n}}{2d} \\[7ex] (10.)\;\; d = \dfrac{(p - a)(p + a)}{2 * SAS_n - p - a} $


$ (1.)\:\: GS_n = ar^{n - 1} \\[5ex] (2.)\:\: SGS_n = \dfrac{a(r^{n} - 1)}{r - 1} \:\:for\:\: r \gt 1 \\[7ex] (3.)\:\: SGS_n = \dfrac{a(1 - r^{n})}{1 - r} \:\:for\:\: r \lt 1 \\[7ex] (4.)\:\: n = \dfrac{\log{\left[\dfrac{SGS_n(r - 1)}{a} + 1\right]}}{\log r} \\[7ex] (5.)\:\: If\:\:r \lt 1,\:\:the\:\:series\:\:converges\:\:and\:\: S_{\infty} = \dfrac{a}{1 - r} \\[7ex] (6.)\:\: If\:\:r \gt 1,\:\:the\:\:series\:\:diverges \\[5ex] (7.)\:\: If\:\:r = 1,\:\:S_{\infty}\:\:DNE \\[5ex] (8.)\:\: r = \dfrac{S_{\infty} - a}{S_{\infty}} \\[7ex] (9.)\:\: a = S_{\infty}(1 - r) $


$ QS = 1st,\:\:\:\:2nd,\:\:\:3rd,\:\:\:4th,... \\[5ex] QS_n = an^2 + bn + c \\[5ex] (1.)\:\: a = \dfrac{1st + 3rd - 2(2nd)}{2} \\[7ex] (2.)\:\: b = \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} \\[7ex] (3.)\:\: c = 3(1st) - 3(2nd) + 3rd \\[5ex] (4.)\:\: \therefore QS_n = \dfrac{1st + 3rd - 2(2nd)}{2} * n^2 + \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} * n + 3(1st) - 3(2nd) + 3rd \\[7ex] The\:\:Left\:\:Hand\:\:Side\:\:must\;\;be\:\:equal\:\:to\;\;the\:\:Right\:\:Hand\:\:Side \\[5ex] (5.)\:\: a + b + c = 1st \\[5ex] (6.)\:\: 4a + 2b + c = 2nd \\[5ex] (7.)\:\: 9a + 3b + c = 3rd \\[5ex] (8.)\:\: 3a + b = 2nd - 1st \\[5ex] (9.)\:\: 8a + 2b = 3rd - 1st $


$ \underline{Triangular\;\;Number\;\;Sequence} \\[3ex] (1.)\:\: TS_n = \dfrac{n(n + 1)}{2} \\[7ex] (2.)\;\; n = \dfrac{\sqrt{8 * TS_n + 1} - 1}{2} \\[7ex] (3.)\:\: TS_n = C(n + 1, 2)...Combinatorics\;\;symbol\;\;for\;\;Combinations \\[5ex] (4.)\;\; STS_n = \dfrac{n(n + 1)(n + 2)}{6} \\[7ex] (5.)\:\: STS_n = C(n + 2, 3)...Combinatorics\;\;symbol\;\;for\;\;Combinations \\[7ex] \underline{Square\;\;Number\;\;Sequence} \\[3ex] (1.)\:\: SS_n = n^2 \\[5ex] (2.)\;\; n = \sqrt{SS_n} \\[5ex] (3.)\;\; SSS_n = \dfrac{n(n + 1)(2n + 1)}{6} \\[7ex] \underline{Cube\;\;Number\;\;Sequence} \\[3ex] (1.)\:\: CS_n = n^3 \\[5ex] (2.)\;\; n = \sqrt[3]{CS_n} \\[5ex] (3.)\;\; SCS_n = \left[\dfrac{n(n + 1)}{2}\right]^2 \\[7ex] (4.)\;\; n = \dfrac{\sqrt{8\sqrt{SCS_n} + 1} - 1}{2} \\[7ex] $


$ \underline{First-Order\;\;Linear\;\;Recurrence\;\;Relation} \\[3ex] (1.)\:\: RS_{n + 1} = r * RS_{n} + a \\[5ex] (2.)\:\: RS_{n + 1} = \dfrac{RS_1 * r^n(r - 1) + a(r^n - 1)}{r - 1} \;\;\;for\;\;r \gt 1 \\[7ex] (3.)\;\; RS_{n + 1} = \dfrac{RS_1 * r^n(1 - r) + a(1 - r^n)}{1 - r} \;\;\;for\;\;r \lt 1 \\[7ex] \underline{Fibonacci\;\;Sequence} \\[3ex] (1.)\;\; \phi = \dfrac{1 + \sqrt{5}}{2} \\[7ex] (2.)\;\; FS_n = \dfrac{\sqrt{5}}{5}\left[\left(\dfrac{1 + \sqrt{5}}{2}\right)^n - \left(\dfrac{1 - \sqrt{5}}{2}\right)^n\right] \\[7ex] (3.)\;\; FS_n = \dfrac{\sqrt{5}}{5}\left[\phi^n - \left(\dfrac{1 - \sqrt{5}}{2}\right)^n\right] \\[7ex] (4.)\;\; SFS_n = \dfrac{\sqrt{5}}{5}\left[\dfrac{\phi^3(\phi^{n - 1} - 1) + [(\phi - 1)(1 - \phi^2)][(1 - \phi)^{n - 1} - 1]}{\phi(\phi - 1)}\right] + 1 $

(1.) How many numbers between 1 and 100 are divisible by 7?


Between 1 and 100
The closest number after 1 which is divisble by 7 is 7
The closest number before 100 which is divisible by 7 is 98
We get the sequence by adding 7 to subsequent terms after 7 until we get to 98
This implies that:
The sequence is an arithmetic sequence
The first term, a = 7
The common difference, d = 7
The last term, p = $AS_n$ = 98
So, the sequence looks like this:
7, 7 + 7 = 14, 14 + 7 = 21, ..., 98
7, 14, 21, ..., 98
Let us find the number of terms, n

$ AS_n = a + d(n - 1) \\[3ex] 98 = 7 + 7(n - 1) \\[3ex] 7 + 7(n - 1) = 98 \\[3ex] 7(n - 1) = 98 - 7 \\[3ex] 7(n - 1) = 91 \\[3ex] n - 1 = \dfrac{91}{7} \\[5ex] n - 1 = 13 \\[3ex] n = 13 + 1 \\[3ex] n = 14 $
(2.) The terms $\dfrac{1}{64}, \dfrac{1}{32}, \dfrac{1}{16}, ...$ from a geometric progression (G.P).

If the sum of the G.P is $2^{36} - 2^{-6}$, find the number of terms.


$ \text{first term} = a \\[3ex] \text{common ratio} = r \\[3ex] \text{number of terms} = n \\[3ex] \text{sum of G.P} = SGP_n \\[5ex] \dfrac{1}{64}, \dfrac{1}{32}, \dfrac{1}{16}, ... \\[5ex] a = \dfrac{1}{64} \\[5ex] r = \dfrac{1}{32} \div \dfrac{1}{64} \\[5ex] r = \dfrac{1}{32} * \dfrac{64}{1} \\[5ex] r = 2 \\[3ex] SGP_n = 2^{36} - 2^{-6} \quad\dots\text{Given} \\[3ex] SGP_n = \dfrac{a(r^{n} - 1)}{r - 1} \:\:for\:\: r \gt 1 \quad\dots\text{Formula} \\[5ex] = a(r^{n} - 1) \div (r - 1) \\[3ex] = \dfrac{1}{64}(2^n - 1) \div (2 - 1) \\[5ex] = \dfrac{2^n - 1}{64} \div 1 \\[5ex] = \dfrac{2^n - 1}{2^6} \\[5ex] \text{Equate both terms: Formula = Given} \\[3ex] \dfrac{2^n - 1}{2^6} = 2^{36} - 2^{-6} \\[5ex] 2^n - 1 = 2^6(2^{36} - 2^{-6}) \\[3ex] 2^n - 1 = 2^{6 + 36} - 2^{6 + - 6} \\[3ex] 2^n - 1 = 2^{42} - 2^0 \\[3ex] 2^n - 1 = 2^{42} - 1 \\[3ex] 2^n = 2^{42} \\[3ex] \text{same base; equate exponents} \\[3ex] n = 42 $
(3.) Can you explain why $\sum\limits_{i = 0}^{n} (n - i) = \dfrac{n(n + 1)}{2}$


$ \sum\limits_{i = 0}^{n} (n - i) \\[5ex] \text{when }: \hspace{3em} \text{term }: \\[3ex] i = 0 \hspace{4em} n - 0 = n \\[3ex] i = 1 \hspace{4em} n - 1 \\[3ex] i = 2 \hspace{4em} n - 2 \\[3ex] i = 3 \hspace{4em} n - 3 \\[3ex] \quad\dots\text{and the series continues}. \\[3ex] \text{The series is:} \\[3ex] n, n - 1, n - 2, n - 3, etc. \\[3ex] \text{This is an Arithmetic Sequence} \\[3ex] SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] \text{where} \\[3ex] SAS_n = \text{sum of the n terms of an arithmetic sequence} \\[3ex] n = \text{number of terms} = n \\[3ex] a = \text{first term} = n \\[3ex] d = \text{common difference} \\[3ex] = (n - 1) - n \\[3ex] = n - 1 - n \\[3ex] = -1 \\[3ex] \implies \\[3ex] SAS_n = \dfrac{n}{2}[2n + -1(n - 1)] \\[5ex] = \dfrac{n}{2}(2n - n + 1) \\[5ex] = \dfrac{n}{2}(n + 1) \\[5ex] = \dfrac{n(n + 1)}{2} $
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