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[∏(3+10n), n=0 to 19] mod 5 =1 Remainder. From Wolfram/Alpha:

http://www.wolframalpha.com/input/?i=%5B%E2%88%8F(3%2B10n),+n%3D0+to+19%5D+mod+5+%3D%3F

9+14+19+80+....+(5n+4)= [n/2] (5n+13)

Show that this is true for  n = 1

(1/2)(5[1] + 13]  =  (1/2)(18)  =  9

Assume that it is true for n =  k.....that is....

9 + 14 + 19 + 80 +........+ (5k + 4)  = (k/2) (5k + 13)      (1)

Prove that it is true for n = k + 1

That is

9 + 14 + 19 + 80 +........+  (5(k + 1) + 4 )  = [ (k + 1)/2 ] [5(k + 1) + 13 ]

Add   (5(k + 1) + 4 )    to both sides of (1)

9 + 14 + 19 + 80 +........+ (5k + 4) + (5(k + 1) + 4 )  =   (k /2)(5k + 13) + (5(k + 1) + 4)

9 + 14 + 19 + 80 +........+ (5k + 4) + (5(k + 1) + 4 )  =   (k /2)(5k + 13) + (5k + 9)

9 + 14 + 19 + 80 +........+ (5k + 4) + (5(k + 1) + 4 )  =    [ (k)(5k + 13) + (10k + 18 ] /2

9 + 14 + 19 + 80 +........+ (5k + 4) + (5(k + 1) + 4 )  =    [ 5k^2 + 13k + 10k + 18] /2

9 + 14 + 19 + 80 +........+ (5k + 4) + (5(k + 1) + 4 )  =    [ 5k^2 + 23k + 18] /2

9 + 14 + 19 + 80 +........+ (5k + 4) + (5(k + 1) + 4 )  =    [ (k + 1) (5k + 18)] /2

9 + 14 + 19 + 80 +........+ (5k + 4) + (5(k + 1) + 4 )  = [ (k + 1)/2] [5(k + 1) + 13 ]

9 + 14 + 19 + 80 +........+  (5(k + 1) + 4 )  = [ (k + 1)/2] [5(k + 1) + 13 ]

Which is what we wished to prove

The sum,S, is given by :

S =   a₁ [ 1 - r^n] / [ 1 - r ]    where a1 = 8, r = 4 and n = 4   ....  so we have

S =  8 [ 1 - 4^4] / [1 - 4]   =  680

The number of possibilites  = 6C3   = 20    ...and we have

{A,B,C}   {A,C,D}  {A,D,E}  {A,E,F}

{A,B,D}   {A,C,E}  {A,D,F}

{A,B,E}   {A,C,F}

{A,B,F}

{B,C,D}  {B,D,E}

{B,C,E}  {B,D,F}

{B,C,F}  {B,E,F}

{C,D,E}

{C,D,F}

{C,E,F}

{D,E,F}

The total is represented by the sum of the first 22 positive integers......the general form is given by:

(n)(n + 1)/ 2    .....in this case n = 22  .....so we have.....

(22)(23) / 2   = 253 logs

[22 x 21] / 2=231 or "c" 

It is "c"

-3, -6, -12, -24.

It is "b"

(3×13×23×33×43×...×183×193) mod 5 =1

[60 x 61] / 2 =1,830 or "e"

Find the 5th term in the binomial expansion.

(a + t) ^17

This is given by:

C (17 , 4) *  a¹³ * t ^t4  =

2380 a¹³ t¹⁴

It is "d" 
 

Find the 7th term in the binomial expansion.

(a − d) ^8

The 7th term is given by

C(8,6) * a² * d^{6  =}  

28a²d⁶  →  " a"

26994.56

Evaluate the expression.

( 94 )

  93 

I assume you want : C(94,93). If so, then:

94! / [94-93]! . 93! =94 or "a"

C (6, 2)=6! / [6-2]!. 2!=15 or "a"

The correct answer is "e" →  3n + 2

3(1) + 2  = 5

3(2) + 2 = 8

3(3) + 2 = 11

3(4) + 2 = 14

3(5) + 2 = 17  ..... !!!!

I'm assuming that "r" must be "1/4" rather than "14"

If so, the terms of the series can be defined as

a_n = 5(1/4)^{n - 1}

So....the first four terms are

5, 5/4, 5/16, 5/64  → Answers  c, e, f and g

It is the "b" sequence.

It is "c" =an=3n

Yes. It is geometric. The common ratio is e=-3

If the j, g and c represent units (joules, grams, celcius?) then the result is 0.4333*2*412.1 j

or 357.126 j

.

3 x $4 =$12 Total cost od the 3 cards

$12 - $6.80=$5.20 The cost of the 3rd card. So that:

$6.80 + $5.20 =$12 /3 =$4 The average of the 3 cards.

Hence the 'averages'  and approximate..