Solve-for-n-i-n-1-n-C-i-2-i-65-n-Z-where-zero-is-included-

Tinku Tara June 4, 2023 Permutation and Combination 0 Comments

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Question Number 57688 by Tawa1 last updated on 10/Apr/19

Solve for n: Σ_i ^(n − 1) ^n C_i 2^i = 65, n ∈ Z^+ . where zero is included

Solve for n : \underset{i}{\sum^{n - 1}} \overset{n}{} C_{i} 2^{i} = 65, n \in Z^{+} . where zero is

included

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Apr/19

question not clear .. Σ_(i=0) ^(n−1) C_i ^n 2^i ←is it the question

q u e s t i o n n o t c l e a r . .

\underset{i = 0}{\sum^{n - 1}} C_{i}^{n} 2^{i} \leftarrow i s i t t h e q u e s t i o n

Commented by Tawa1 last updated on 10/Apr/19

Yes sir.

Yes sir .

Commented by Abdo msup. last updated on 10/Apr/19

first we have Σ_(i=0) ^(n−1) C_n ^i 2^i =Σ_(i=0) ^n C_n ^i 2^i −C_n ^n 2^n =3^n −2^n (we suppose that the sum begins from 0) ⇒(e) ⇒3^n −2^n =65 ⇒ n=4 .

f i r s t w e h a v e \sum_{i = 0}^{n - 1} C_{n}^{i} 2^{i} = \sum_{i = 0}^{n} C_{n}^{i} 2^{i} - C_{n}^{n} 2^{n}

= 3^{n} - 2^{n} (w e s u p p o s e t h a t t h e s u m b e g i n s f r o m 0)

\Rightarrow (e) \Rightarrow 3^{n} - 2^{n} = 65 \Rightarrow n = 4 .

Commented by Tawa1 last updated on 10/Apr/19

God bless you sirs, is there anyway to find n from 3^n − 2^n = 65

God bless you sirs, is there anyway to find n from 3^{n} - 2^{n} = 65

Commented by Tawa1 last updated on 10/Apr/19

Sir, please i need the rules of summation, expecially the sum that has something like Σ_(i = 0) ^(n − 1) how to split them

Sir, please i need the rules of summation, expecially the sum

that has something like \underset{i = 0}{\sum^{n - 1}} how to split them

Commented by Tawa1 last updated on 10/Apr/19

Because have been trying to understand how to get 3^n − 2^n = 65. I did not understand yet.

Because have been trying to understand how to get

3^{n} - 2^{n} = 65 . I did not understand yet .

Commented by Tawa1 last updated on 10/Apr/19

Sorry for disturbing sir

Sorry for disturbing sir

Commented by Tawa1 last updated on 10/Apr/19

or textbook i can learn it.

or textbook i can learn it .

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Apr/19

S=C_0 ^n 2^0 +C_1 ^n 2^1 +C_2 ^n 2^2 +....+C_(n−1) ^n 2^(n−1) now look here (x+2)^n =C_0 ^n x^(n−0) 2^0 +C_1 ^n x^(n−1) 2^1 +C_2 ^n x^(n−2) 2^2 +...+C_(n−1) ^n x^1 2^(n−1) +C_n ^n x^(n−n) 2^n now put x=1 both side 3^n =[C_0 ^n 2^0 +C_1 ^n 2^1 +...+C_(n−1) ^n 2^(n−1) ]+C_n ^n 2^n ←look here 3^n =S+C_n ^n 2^n now C_n ^n =((n!)/(n!(n−n)!))=1 so 3^n =S+2^n ×1 S=3^n −2^n so Σ_(i=0) ^(n−1) C_i ^n_ 2^i =S=3^n −2^n 3^n −2^n =65 by lnspection 3^n −2^n =81−16 3^n −2^n =3^4 −2^4 →n=4

S = C_{0}^{n} 2^{0} + C_{1}^{n} 2^{1} + C_{2}^{n} 2^{2} + \dots . + C_{n - 1}^{n} 2^{n - 1}

n o w l o o k h e r e

{(x + 2)}^{n} = C_{0}^{n} x^{n - 0} 2^{0} + C_{1}^{n} x^{n - 1} 2^{1} + C_{2}^{n} x^{n - 2} 2^{2} + \dots + C_{n - 1}^{n} x^{1} 2^{n - 1} + C_{n}^{n} x^{n - n} 2^{n}

n o w p u t x = 1 b o t h s i d e

3^{n} = [C_{0}^{n} 2^{0} + C_{1}^{n} 2^{1} + \dots + C_{n - 1}^{n} 2^{n - 1}] + C_{n}^{n} 2^{n} \leftarrow l o o k h e r e

3^{n} = S + C_{n}^{n} 2^{n}

n o w C_{n}^{n} = \frac{n!}{n! (n - n)!} = 1

s o

3^{n} = S + 2^{n} \times 1

S = 3^{n} - 2^{n}

s o \underset{i = 0}{\sum^{n - 1}} C_{i}^{n_{}} 2^{i} = S = 3^{n} - 2^{n}

3^{n} - 2^{n} = 65

b y l n s p e c t i o n

3^{n} - 2^{n} = 81 - 16

3^{n} - 2^{n} = 3^{4} - 2^{4} \to n = 4

Commented by Tawa1 last updated on 10/Apr/19

Ohh, I appreciate your time and effort sir. God bless you sir

Ohh, I appreciate your time and effort sir . God bless you sir

Commented by Tawa1 last updated on 10/Apr/19

I really appreciate sir

I really appreciate sir

Commented by Tawa1 last updated on 10/Apr/19

And i understand very well

And i understand very well

Answered by mr W last updated on 10/Apr/19

assumed you mean Σ_(i=0) ^(n − 1) ^n C_i 2^i = 65 LHS=(1+2)^n −2^n (1+2)^n −2^n =65 3^n −2^n =65 ⇒n=4

a s s u m e d y o u m e a n

\underset{i = 0}{\sum^{n - 1}} \overset{n}{} C_{i} 2^{i} = 65

L H S = {(1 + 2)}^{n} - 2^{n}

{(1 + 2)}^{n} - 2^{n} = 65

3^{n} - 2^{n} = 65

\Rightarrow n = 4

Commented by Tawa1 last updated on 10/Apr/19

How is left hand side = (1 + 2)^n − 2^n sir ??

How is left hand side = {(1 + 2)}^{n} - 2^{n} sir ? ?

Commented by mr W last updated on 10/Apr/19

(1+2)^n =Σ_(i=0) ^n C_i ^n 2^i =Σ_(i=0) ^(n−1) C_i ^n 2^i +2^n =LHS+2^n

{(1 + 2)}^{n} = \underset{i = 0}{\sum^{n}} C_{i}^{n} 2^{i} = \underset{i = 0}{\sum^{n - 1}} C_{i}^{n} 2^{i} + 2^{n} = L H S + 2^{n}

Commented by Tawa1 last updated on 10/Apr/19

God bless you sir. any step to get n = 4 ?

God bless you sir . any step to get n = 4 ?

Commented by mr W last updated on 10/Apr/19

try and see...

t r y a n d s e e \dots

Commented by Tawa1 last updated on 10/Apr/19

Trier and error ?

Trier and error ?

Commented by Tawa1 last updated on 10/Apr/19

I found a method i used to get n = 4, from 3^n − 2^n = 65. I used algebra representation. But am still finding it difficult to understand. how to get 3^n − 2^n = 65 from the relation.

I found a method i used to get n = 4, from 3^{n} - 2^{n} = 65 . I used algebra

representation .

But am still finding it difficult to understand . how to get

3^{n} - 2^{n} = 65 from the relation .

Commented by Tawa1 last updated on 10/Apr/19

Maybe i need sigma notation rules. How to apply

Maybe i need sigma notation rules . How to apply

Commented by Tawa1 last updated on 10/Apr/19

3^n − 2^n = 65 [3^((n/2)) ]^2 − [2^((n/2)) ]^2 = 65 3^(n/2) = m and 2^(n/2) = y m^2 − y^2 = 65 (m + y)(m − y) = 13 × 5 m + y = 13 m − y = 5 m = 9, y = 4 3^(n/2) = 3^2 and 2^(n/2) = 2^2 n = 4 twice.

3^{n} - 2^{n} = 65

{[3^{(n / 2)}]}^{2} - {[2^{(n / 2)}]}^{2} = 65

3^{n / 2} = m and 2^{n / 2} = y

m^{2} - y^{2} = 65

(m + y) (m - y) = 13 \times 5

m + y = 13

m - y = 5

m = 9, y = 4

3^{n / 2} = 3^{2} and 2^{n / 2} = 2^{2}

n = 4 twice .

Commented by Tawa1 last updated on 10/Apr/19

But am still comfused on how to get 3^n − 2^n from the sum

But am still comfused on how to get 3^{n} - 2^{n} from the sum

Commented by MJS last updated on 10/Apr/19

this isn′t a method, it works here just by chance 3^n −2^n =91 (3^(n/2) )^2 −(2^(n/2) )^2 =91 (3^(n/2) −2^(n/2) )(3^(n/2) +2^(n/2) )=7×13 3^(n/2) −2^(n/2) =7 3^(n/2) +2^(n/2) =13 ⇒ 3^(n/2) =10 ∧ 2^(n/2) =3 ⇒ n=((2ln 10)/(ln 3)) ∧ n=((2ln 3)/(ln 2)) while in this case n≈4.28235

this {isn}^{'} t a method, it works here just by chance

3^{n} - 2^{n} = 91

{(3^{n / 2})}^{2} - {(2^{n / 2})}^{2} = 91

(3^{n / 2} - 2^{n / 2}) (3^{n / 2} + 2^{n / 2}) = 7 \times 13

3^{n / 2} - 2^{n / 2} = 7

3^{n / 2} + 2^{n / 2} = 13

\Rightarrow 3^{n / 2} = 10 \land 2^{n / 2} = 3

\Rightarrow n = \frac{2 \ln 10}{\ln 3} \land n = \frac{2 \ln 3}{\ln 2}

while in this case n \approx 4.28235

Commented by Tawa1 last updated on 10/Apr/19

Thank you sir.

Thank you sir .

Commented by Tawa1 last updated on 10/Apr/19

If i really understand the spliting to 3^n − 2^n . it will help. Am sorry for disturbance sir

If i really understand the spliting to 3^{n} - 2^{n} . it will help .

Am sorry for disturbance sir

Commented by Tawa1 last updated on 10/Apr/19

I don′t want to cram

I {don}^{'} t want to cram

Commented by mr W last updated on 10/Apr/19

for how to solve equations like 3^n −2^n =65 there is only one way: that′s try and error. since n≥1, you try with n=1, 3^1 −2^1 =1<65⇒not good, then try with n=2, 3^2 −2^2 =5<65⇒not good, then try with n=3, 3^3 −2^3 =19<65⇒not good, then try with n=4, 3^4 −2^4 =65=65⇒good, you got it! to be sure you try with n=5, 3^5 −2^5 =211>65⇒not good, no other solution.

f o r h o w t o s o l v e e q u a t i o n s l i k e

3^{n} - 2^{n} = 65

t h e r e i s o n l y o n e w a y : {t h a t}^{'} s t r y a n d e r r o r .

s i n c e n ⩾ 1, y o u t r y w i t h n = 1,

3^{1} - 2^{1} = 1 < 65 \Rightarrow n o t g o o d, t h e n t r y w i t h n = 2,

3^{2} - 2^{2} = 5 < 65 \Rightarrow n o t g o o d, t h e n t r y w i t h n = 3,

3^{3} - 2^{3} = 19 < 65 \Rightarrow n o t g o o d, t h e n t r y w i t h n = 4,

3^{4} - 2^{4} = 65 = 65 \Rightarrow g o o d, y o u g o t i t!

t o b e s u r e y o u t r y w i t h n = 5,

3^{5} - 2^{5} = 211 > 65 \Rightarrow n o t g o o d, n o o t h e r s o l u t i o n .

Commented by Tawa1 last updated on 10/Apr/19

Ohh, God bless you sir. I appreciate your time sir.

Ohh, God bless you sir . I appreciate your time sir .

Commented by Tawa1 last updated on 10/Apr/19

When you are chance sir, i need the general term of Σ_(i = 0) ^(n − 1) ...

When you are chance sir, i need the general term of \underset{i = 0}{\sum^{n - 1}} \dots

Commented by mr W last updated on 10/Apr/19

what′s your problem? (1) you don′t understand how to get 3^n −2^n =65 from Σ_(i=0) ^(n − 1) ^n C_i 2^i = 65. (2) you don′t understand how to get n=4 from 3^n −2^n =65.

{w h a t}^{'} s y o u r p r o b l e m ?

(1) y o u {d o n}^{'} t u n d e r s t a n d h o w t o g e t

3^{n} - 2^{n} = 65 f r o m \underset{i = 0}{\sum^{n - 1}} \overset{n}{} C_{i} 2^{i} = 65 .

(2) y o u {d o n}^{'} t u n d e r s t a n d h o w t o g e t

n = 4 f r o m 3^{n} - 2^{n} = 65 .

Commented by Tawa1 last updated on 10/Apr/19

(1) is my problem sir.

(1) is my problem sir .

Commented by mr W last updated on 10/Apr/19

i can only give you examples, try to find the rules by yourself: a_1 +a_2 +a_3 +...+a_n =Σ_(i=1) ^n a_i a_1 +a_2 +a_3 +...+a_n =a_1 +Σ_(i=2) ^n a_i a_1 +a_2 +a_3 +...+a_(n−1) +a_n =Σ_(i=1) ^(n−1) a_i +a_n a_1 +a_2 +a_3 +...+a_(n−2) +a_(n−1) +a_n =Σ_(i=2) ^(n−2) a_i +a_1 +a_(n−1) +a_n

i c a n o n l y g i v e y o u e x a m p l e s, t r y t o f i n d

t h e r u l e s b y y o u r s e l f :

a_{1} + a_{2} + a_{3} + \dots + a_{n} = \underset{i = 1}{\sum^{n}} a_{i}

a_{1} + a_{2} + a_{3} + \dots + a_{n} = a_{1} + \underset{i = 2}{\sum^{n}} a_{i}

a_{1} + a_{2} + a_{3} + \dots + a_{n - 1} + a_{n} = \underset{i = 1}{\sum^{n - 1}} a_{i} + a_{n}

a_{1} + a_{2} + a_{3} + \dots + a_{n - 2} + a_{n - 1} + a_{n} = \underset{i = 2}{\sum^{n - 2}} a_{i} + a_{1} + a_{n - 1} + a_{n}

Commented by mr W last updated on 10/Apr/19

3^n =(1+2)^n =Σ_(k=0) ^n C_k ^( n) 2^k =Σ_(k=0) ^(n−1) C_k ^( n) 2^k +C_n ^( n) 2^n is this clear for you?

3^{n} = {(1 + 2)}^{n} = \underset{k = 0}{\sum^{n}} C_{k}^{n} 2^{k} = \underset{k = 0}{\sum^{n - 1}} C_{k}^{n} 2^{k} + C_{n}^{n} 2^{n}

i s t h i s c l e a r f o r y o u ?

Commented by Tawa1 last updated on 10/Apr/19

Yes sir, very clear sir. I realy appreciate your time sir. God bless you sir

Yes sir, very clear sir . I realy appreciate your time sir . God bless you sir

Commented by mr W last updated on 10/Apr/19

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