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Assignment 4 Of MTH202 solution Fall 2010
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#21
Bhai Xpert......??? Kindly.... MTH202 Assignment no 4 solution Upload ker dooo ........................ only 2 hours left
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#22
Any one one have any question solved ?
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#23
i will be uploading if WAPDA allows :p
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#24
plz upload kar do solution
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#25
Yaar ker bhe doo .....
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#26
using mathematical induction

step 1: first we will prove that for n =1 it is true

thus x^2n-1 = x^2-1 = (x-1)*(x+1)

clearly since it has factor of x+1 we can say that x^2n-1 is divisible by x-1

step 2: let assume that for n = k it is true thus

x^2k-1 is divisible by x+1

thus we can write as x^2k-1 = P (x+1) P is quotient

now we have to prove that it is true for n=k+1

step 3: now let n = k+1 thus

x^2(k+1) - 1 = x^(2k+2) - 1 = x^2k*x^2 -1 = (x^2k-1+1)x^2 - 1

(x^2k-1)*x^2 + (x^2-1)

for step 2 we can write above equation as

P (x+1)*x^2 + (x+1)*(x-1) = (x+1)* (Px^2+x-1)

which contain factor of x+1 thus divisible by x+1

thus even it is proved for n=k+1

according to mathematical induction given is true for all integral values of n


Question 3; Marks: 03
Use the Euclidean algorithm to find gcd (1331, 1001)
Solution
gcd(1331, 1001) = gcd(1001, 330)
= gcd(330, 11)
= gcd(11, 0)
= 11


How many bit strings of length 10 have

a) Exactly three 0’s?

is ka answer hai 10C3 =120

b) The same number of 0’s as 1’s?
is ka answer hai 10 C 5 = 152
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#27
Dear students
Main nay yeh subject nahi perha howa.warna aapko aik hi solution provide kerta 100 % accurate khud solve ker kay.

Magar mujhay dosray dosto say yeh 2 solution milay hai .aap yeh subject per rahay hoo aapko zayada behter pata hoga k khon sa theek hai.
Ajj grace day hai main nay kaha yeh na hoo aapki assignment submit honay say reh jaye is liye upload ker diye hai solution.
so please check the attachments and submit the correct one
thanks


Attached Files
.doc   idea solution 1.doc (Size: 27 KB / Downloads: 133)
.doc   idea solution 2.doc (Size: 29.5 KB / Downloads: 136)
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#28
dear sir
today is margin day of mth 202 assigment # 4 plz upload the solution
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