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Question Number 120636 by TANMAY PANACEA last updated on 01/Nov/20

selective Binomial

s e l e c t i v e B i n o m i a l

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by TANMAY PANACEA last updated on 01/Nov/20

Commented by Dwaipayan Shikari last updated on 01/Nov/20

S=(1/(1!(n−1)!))+(1/(3!(n−3)!))+... S=(n/(n!))+((n(n−1)(n−2))/(3!n!))+((n(n−1)(n−2)(n−3)(n−4))/(5!n!))+... =(1/(n!))(n+((n(n−1)(n−2))/(3!))+((n(n−1)(n−2)(n−3)(n−4))/(5!))+...) =(2^(n−1) /(n!)) (1+x)^n =1+nx+((n(n−1))/(2!))x^2 +... (1−x)^n =1−nx+((n(n−1))/(2!))x^2 −.. (((1+x)^n −(1−x)^n )/2)=nx+((n(n−1)(n−2))/(3!))x^3 +... 2^(n−1) =(n+((n(n−1)(n−2))/(3!))+...) (x=1)

S = \frac{1}{1! (n - 1)!} + \frac{1}{3! (n - 3)!} + \dots

S = \frac{n}{n!} + \frac{n (n - 1) (n - 2)}{3! n!} + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{5! n!} + \dots

= \frac{1}{n!} (n + \frac{n (n - 1) (n - 2)}{3!} + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{5!} + \dots)

= \frac{2^{n - 1}}{n!}

{(1 + x)}^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \dots

{(1 - x)}^{n} = 1 - n x + \frac{n (n - 1)}{2!} x^{2} - . .

\frac{{(1 + x)}^{n} - {(1 - x)}^{n}}{2} = n x + \frac{n (n - 1) (n - 2)}{3!} x^{3} + \dots

2^{n - 1} = (n + \frac{n (n - 1) (n - 2)}{3!} + \dots) (x = 1)

Commented by Dwaipayan Shikari last updated on 01/Nov/20

(1+x)^n =C_0 +C_1 x+... ∫_0 ^1 (1+x)^n =∫_0 ^1 C_0 +C_1 x+.. (2^(n+1) /(n+1))=C_0 +(1/2)C_1 +(1/3)C_2 +...(1/(n+1))C_n

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

\int_{0}^{1} {(1 + x)}^{n} = \int_{0}^{1} C_{0} + C_{1} x + . .

\frac{2^{n + 1}}{n + 1} = C_{0} + \frac{1}{2} C_{1} + \frac{1}{3} C_{2} + \dots \frac{1}{n + 1} C_{n}

Commented by Dwaipayan Shikari last updated on 01/Nov/20

C_0 +(C_0 +C_1 )+(C_0 +C_1 +C_2 )+... n(C_0 +C_1 +C_2 +...)−(C_1 +2C_2 +3C_3 +..) n.2^n −n.2^(n−1) =n2^(n−1) (2−1)=n.2^(n−1) [(1+x)^n =C_0 +C_1 x+C_3 x^2 +.. n(1+x)^(n−1) =C_1 +2C_3 x+.. n.2^(n−1) =C_1 +2C_3 +...]

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + \dots

n (C_{0} + C_{1} + C_{2} + \dots) - (C_{1} + 2 C_{2} + 3 C_{3} + . .)

n . 2^{n} - n . 2^{n - 1} = n 2^{n - 1} (2 - 1) = n . 2^{n - 1}

[{(1 + x)}^{n} = C_{0} + C_{1} x + C_{3} x^{2} + . .

n {(1 + x)}^{n - 1} = C_{1} + 2 C_{3} x + . .

n . 2^{n - 1} = C_{1} + 2 C_{3} + \dots]

Commented by Dwaipayan Shikari last updated on 01/Nov/20

(1+x+2x^2 )^(20) =a_0 +a_1 x+... 1^(20) =a_0 (x=0) (2)^(40) =1+a_1 +a_2 +.... (x=1) 2^(20) =1−a_1 +a_2 −a_3 +....(x=−1) 2^(40) +2^(20) =2(1+a_2 +a_4 +...+a_(38) ) 2^(39) +2^(19) =a_0 +a_2 +.... 2^(19) (2^(20) +1)=a_0 +a_2 +...+a_(38)

{(1 + x + 2 x^{2})}^{20} = a_{0} + a_{1} x + \dots

1^{20} = a_{0} (x = 0)

{(2)}^{40} = 1 + a_{1} + a_{2} + \dots . (x = 1)

2^{20} = 1 - a_{1} + a_{2} - a_{3} + \dots . (x = - 1)

2^{40} + 2^{20} = 2 (1 + a_{2} + a_{4} + \dots + a_{38})

2^{39} + 2^{19} = a_{0} + a_{2} + \dots .

2^{19} (2^{20} + 1) = a_{0} + a_{2} + \dots + a_{38}

Commented by TANMAY PANACEA last updated on 01/Nov/20

thank you

t h a n k y o u

Commented by TANMAY PANACEA last updated on 01/Nov/20

i have posted questions collected from[books of IIT

i h a v e p o s t e d q u e s t i o n s c o l l e c t e d f r o m [b o o k s o f I I T

Commented by Dwaipayan Shikari last updated on 01/Nov/20

I am in class 11 (high school) :)

I a m i n c l a s s 11 (h i g h s c h o o l) :)

Commented by prakash jain last updated on 02/Nov/20

You are doing very well for class XI. Are you aiming for IIT?

You are doing very well for class XI .

Are you aiming for IIT ?

Commented by Dwaipayan Shikari last updated on 02/Nov/20

I want to study physics in later. IIT′s are good but i want to study freely

I w a n t t o s t u d y p h y s i c s i n l a t e r .

{I I T}^{'} s a r e g o o d b u t i w a n t t o s t u d y f r e e l y

Commented by Dwaipayan Shikari last updated on 02/Nov/20

But in2022 i have to take this exam :)

B u t i n 2022 i h a v e t o t a k e t h i s e x a m :)

Commented by prakash jain last updated on 02/Nov/20

Are u interested in theoritical physics?

Are u interested in theoritical physics ?

Commented by TANMAY PANACEA last updated on 02/Nov/20

17^(1983) +11^(1983) −7^(1983) answer is =1 explanation 7^1 =last digit 7 7^2 =l.d 9 7^3 =l.d 3 7^4 =1 7^5 =7 cycle of (7 9 1 7) ((1983)/7)=283+(2/7)→last digit=9 last digit of 7^(1983) →=9 last digit of 11^(1983) is1 so (9+1−9)→last digit=1 pls chk

17^{1983} + 11^{1983} - 7^{1983}

a n s w e r i s = 1

e x p l a n a t i o n

7^{1} = l a s t d i g i t 7

7^{2} = l . d 9

7^{3} = l . d 3

7^{4} = 1

7^{5} = 7

c y c l e o f (7 9 1 7)

\frac{1983}{7} = 283 + \frac{2}{7} \to l a s t d i g i t = 9

l a s t d i g i t o f 7^{1983} \to= 9

l a s t d i g i t o f 11^{1983} i s 1

s o (9 + 1 - 9) \to l a s t d i g i t = 1

p l s c h k

Commented by TANMAY PANACEA last updated on 02/Nov/20

32^(32^(32) ) devided by 7 (4×7+5)^(32^(32) ) =f(7)+5^(32^(32) ) now 32^(32) =even number 32^(32) 2^1 →2 2^2 →4 2^3 →8 2^4 →16 2^5 →32 cycle(last digit 2 4 8 6 ) 32^(32) =2k is a even number whose last digit 6 5^(2k) =last two digit is 25 so when devided by 7→last digit is 4 pls chk is my thinking correct

32^{32^{32}} d e v i d e d b y 7

{(4 \times 7 + 5)}^{32^{32}}

= f (7) + 5^{32^{32}}

n o w 32^{32} = e v e n n u m b e r

32^{32}

2^{1} \to 2

2^{2} \to 4

2^{3} \to 8

2^{4} \to 16

2^{5} \to 32

c y c l e (l a s t d i g i t 2 4 8 6)

32^{32} = 2 k i s a e v e n n u m b e r w h o s e l a s t d i g i t 6

5^{2 k} = l a s t t w o d i g i t i s 25

s o w h e n d e v i d e d b y 7 \to l a s t d i g i t i s 4

p l s c h k i s m y t h i n k i n g c o r r e c t

Commented by Dwaipayan Shikari last updated on 02/Nov/20

yes , Prakash sir!

y e s, P r a k a s h s i r!

Commented by Dwaipayan Shikari last updated on 03/Nov/20

(4×7+4)=32^(32^(32) ) typo

(4 \times 7 + 4) = 32^{32^{32}}

t y p o

Commented by TANMAY PANACEA last updated on 03/Nov/20

yes yes

y e s y e s

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