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1)The first term of a sequence is 13. Starting with the second term, each term is the sum of the cubes of the digits in the previous term. For example, the second term is \(1^3+3^3=28\) Find the 100th term.

 

2)In a sequence of ten terms, each term (starting with the third term) is equal to the sum of the two previous terms. The seventh term is equal to 6. Find the sum of all ten terms.

 

3)Find \( 100^2 - 99^2 + 98^2 - 97^2 + 96^2 - 95^2 + \dots + 2^2 - 1^2. \)

 Mar 9, 2019
 #1
avatar+7761 
+1

1.

\(a_1 = 13\\ a_2 = 28\\ a_3 = 520\\ a_4 = 133\\ a_5 = 55\\ a_6 = 250\\ a_7 = 133\\ a_8 = 55\\ a_9 = 250\\ \vdots\)

As you see, for all \(n \geq 4\)\(a_n = \begin{cases}133&&n\equiv1\pmod3\\ 55&& n\equiv 2 \pmod 3\\ 250&& n\equiv 0 \pmod 3\end{cases}\).

\(\therefore a_{100} = 133\)

.
 Mar 9, 2019
edited by MaxWong  Mar 9, 2019
 #2
avatar+7761 
+1

2.

Assume \(a_1 = x,\quad a_2 = y\).

\(a_3 = x + y\\ a_4 = x+2y\\ a_5 = 2x+3y\\ a_6 = 3x + 5y\\ a_7 = 5x + 8y\\ a_8 = 8x + 13y\\ a_9 = 13x + 21y\\ a_{10} =21x+34y\)

WLOG, assume y = 0 \(\implies x = \dfrac{6}{5}\).

Sum of 10 terms = \(\dfrac{6}{5} \left(1+0+1+1+2+3+5+8+13+21\right) = \dfrac{6}{5} \cdot(55) = 66\)

.
 Mar 9, 2019
 #3
avatar+7761 
0

For any adjacent terms (2n)2 and (2n - 1)2, difference = 4n - 1.

\(100^2 - 99^2 + 98^2 - 97^2+\cdots + 2^2 - 1^2\\ = 199 + 195 + 191+\cdots + 3\\ =\dfrac{(3+199)(\frac{199-3}{4}+1)}{2}\\ =5050 \)

.
 Mar 9, 2019
edited by MaxWong  Mar 9, 2019
 #4
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0

Max: The last one forms an Arithmetic Sequence as follows:

199, 195, 191, 187, 183......and so on for 50 terms

 

Sum =50/2*[2*199 + (50 -1) * - 4]

Sum =5050

 Mar 9, 2019
 #5
avatar+7761 
+1

Oops. Thank you for the correction. 

MaxWong  Mar 9, 2019
 #6
avatar+105464 
+1

 2)In a sequence of ten terms, each term (starting with the third term) is equal to the sum of the two previous terms. The seventh term is equal to 6. Find the sum of all ten terms.

 

a1 = a1

a2 = a2

a3 = a1 + a2

a4 = a2 + a3    =  a1 + 2a2 

a5 = a3 + a4    =   2a1 + 3a2

a6 =  a4 + a5    = 3a1 + 5a2

a7 =  a5 + a6    =  5a1 + 8a2  = 6    →   a1  =  (1/5) (6 - 8a2)   (1)

a8 =  8a1 + 13a2

a9 = 13a1 + 21a2

10 = 21a1 + 34a2

 

So....the sum of the first ten terms =   55a + 88a2

 

Subbing (1) into this.....the sum is

 

55(1/5) (6 - 8a2) + 88a2  =

 

11(6 - 8a2) + 88a2  =

 

66 - 88a2 + 88a2 =

 

66

 

 

cool cool  cool

 Mar 9, 2019

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