Question

For the given statement Pn, write the statements P1, Pk, and Pk+1.

2 + 4 + 8 + . . . + 2n = 2^(n+1) - 2

Answer 1

p1 mean n=1
same way, pk would mean setting n=k
same for k+1 ..

Answer 2

@shubhamsrg
so p1= n which =1
and pk = n?
so then pk+1 = n+1??
that's a little confusing..

Answer 3

hmm,,yep you are confused..
p denotes the statement ..
n denotes term number
question asks you p1 i.e. statement or expression when n =1
same way, pk would be mean when n= k ..
pn = k has no meaning for this context..

Answer 4

@shubhamsrg do you think you could use them in a simple example?

Answer 5

leme try

let pn be
1 + 2 +3 + 4 .... n

p1 would mean when n= 1 ,that is 1st term
here, 1st term is 1 ,, so p1 =1

pk would mean kth term

nth term is ofcorse = n
so kth term will be likewise =k
so pk =k

for pk+1 ,,just substitute n=k+1 in the nth term..

Answer 6

@shubhamsrg so for my example,
p1= 2
pk = 2n
and pk+1 =( 2(n+1) - 2) +1 or 2n+1 ?

Answer 7

for pk, just substitute k in place of n
for pk+1 .
just substitute k+1 in place of n
p1 is correct though..

Answer 8

so your pk = k

Answer 9

tell me pk+1 now..

Answer 10

@shubhamsrg k+1 ?

Answer 11

@shubhamsrg can i change all the n's in my problem to k's?

Answer 12

yes you can,,without the loss of generality..

Answer 13

@shubhamsrg but what i don't understand is do i have to use the equation on the right? and am i actually plugging in any numbers or anything?

Answer 14

ohh,,didnt see a RHS there,., :P

yes, you are plugging in numbers, natural numbers, in place of n
n denotes the term no.
like 2 is the first term
4 is the second term.
6 is the 3rd term.
and so on
so nth term = 2*n

thus, you can say, kth term = 2*k
k+1 th term = 2(k+1)

same way, do it on right hand side,.

Answer 15

@shubhamsrg so then when you have \[2\times k\] it'll equal \[2\] and when you have \[2k+1 \] it'll equal 3 if \[k=1\]

Answer 16

P\(\color {red} 1\): \(2 = 2(\color{red} 1 + 1) - 2\)
P\(\color{red}k\): \(2 + 4 + 8 + \cdots + 2\color{red}k = 2(\color{red} k + 1) - 2\)

Answer 17

@sirm3d which equation do i plug it into?

Answer 18

the equation Pn: \(\displaystyle 2 + 4 +8 + \cdots + 2n = 2(n+1) -2\)
the problem asks you to write the correct equation when (i) n = 1, (ii) n = k, and (iii) n = k+1

Answer 19

P1 means replace \(n\) in Pn by 1 ...

Answer 20

@sirm3d do i use 2n or 2^(n+1)−2

Answer 21

i am just following Pn from the first post, although i think \[\Large Pn: 2 + 4 + 8 + \cdots +2^n=2^{n+1}-2\] is what you intended to write.

Answer 22

@sirm3d yes , that is what i meant to write. oops.

Answer 23

the statement \(Pn\) says that there are several terms to add in the left hand side of the equation, the first three terms of which are 2, 4 and 8, while the right hand side is an expression involving \(n\)

Answer 24

@sirm3d okay i understand that

Answer 25

so we are cleared with P1 and Pk. can you write Pk+1?

Answer 26

@sirm3d do p1 and pk both equal 2?

Answer 27

nope. P1 is specific, n = 1, while Pk is more general, n= k.

Answer 28

\[\Large P \color{red}n: 2 + 4 + 8 + \cdots + 2^\color{red}n = 2^\color{red}{n+1} -2 \]\[\Large P \color{red}k: 2 + 4 + 8 + \cdots + 2^\color{red}k = 2^\color{red}{k+1} -2 \]

Answer 29

Pk is no different in form from Pn.
P1 on the other hand, is more specific. The number of terms to write on the LHS is controlled by (n=1)

Answer 30

@sirm3d so for p1, you would plug in 1 into the equations?

Answer 31

yes.

Answer 32

@sirm3d so then when you have Pk, how would you do that? :(

Answer 33

replace n by k thoughout the equation

Answer 34

@sirm3d so then it would be \[2k=2^{k=1}-2\]

Answer 35

nope. the tricky part is the left hand side of the equation, 2 + 4 + 8 + ... + 2^n

Answer 36

to write Pk, copy and paste Pn, replace every instance of n by k.

Answer 37

Pn: 2 + 4 + 8 + ... + 2^n = 2^(n+1) - 2
replace n by k in the equation to get you Pk

Answer 38

don't forget that the left hand side of Pn has "many" terms, the last of which is \(\Large \displaystyle 2^n\)

Answer 39

@sirm3d i honestly do not understand this one bit :( i'm sorry :(

Answer 40

@sirm3d it's not that you're explaining it bad or anything, i just don't understand the concepts

Answer 41

let's try by writing P1, P2 P3 and P4 statements

Answer 42

can write them out?

Answer 43

one at a time.

Answer 44

@sirm3d p1= 2
because 2x1 =2

Answer 45

you should write an equation. the left hand side consists of summands, the right hand side an expression
i'll do P1
P1: 2 + 4 + 8 + ... + 2^1 = 2^(1+1) - 2

Answer 46

but, wait, the last term in the LHS is 2, so there's only one term on that side
P1: 2 = 2^(1 + 1) - 2

Answer 47

@sirm3d so the equation that i was given has the left hand side as 2n but if you do the math is should be 2^n. hmm
p2= \[2^2 = 4 \] = \[2^{(2+1)}-2 = 2^3-2 =8-2=6\]
but that is incorrect...

Answer 48

here's what you need to remember in this Pn: 2 + 4 + 8 + ... + 2^n = 2^(n+1)-2
there are probably many terms on the left hand side. the first term is 2, and the last term is 2^n.
so when n = 1, the first term is 2 and the last term is also 2, there's only one term.
now, when n = 2, the first term is 2 and the last term is 2^2=4, so there are two terms.

Answer 49

\[P1: 2 = 2^(1+1) -2\]\[P2:2+4=2^(2+1)-2\]
it would be a wise guess that in P3, there are three terms to add in the LHS

Answer 50

@sirm3d but why are there more terms being added to the left side ?

Answer 51

that is what Pn asserts: there are terms to add on the LHS, while the RHS gives you a quick way of finding the sum without actually adding them.

Answer 52

the LHS of Pn grows as n increases.
P1: 2
P2: 2 + 4=(2^2)
P3 : 2 + 4 + 8(=2^3)
P4: 2 + 4 + 8 + 16(=2^4)
P5: 2 + 4 + 8 + 16 + 32(=2^5)
P10: 2+4+8+16+32+64+128+256+512+1024(=2^10)

Answer 53

the RHS of Pn on the other hand, is just an expression
P1: 2^(1+1) -2
P2: 2^(2+1) -2
P3: 2^(3+1) -2
P4: 2^(4+1) -2
P5: 2^(5+1) -2
P10: 2^(10+1)-2

Answer 54

\[\Large Pn: \underbrace{2+4+8+\cdots+2^n}_{n\text{ terms}}=\underbrace{2^{n+1}-2}_{\text{one expression}}\]

Answer 55

\[\Large P_\color{red}1:\underbrace{?}_{1\text{ term}}=2^{\color{red}1+1}-2\]\[\Large P_\color{red}2:\underbrace{?+?}_{2\text{ terms}}=2^{\color{red}2+1}-2\]\[\Large P_\color{red}3:\underbrace{?+?+?}_{3\text{ terms}}=2^{\color{red}3+1}-2\]\[\Large P_\color{red}4:\underbrace{?+?+?+?}_{4\text{ terms}}=2^{\color{red}4+1}-2\]

Answer 56

@sirm3d ohhh i understand. omg i get it now. that makes a lot more sense. you add the number from the previous P

Answer 57

right.

Answer 58

@sirm3d so i understand it now. thank you soooooo much for your help!! i honestly appreciate it! have a happy new year :)

Answer 59

happy new year to you too.
here's a problem if you wish to practice.
\[\Large \color{red}{Pn}: \color{blue}{2+8+14+\cdots+(6n-4)=3n^2-n}\]