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Notes On Topics Of Algebra

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Attached files contain the basic concept of Algebra topics.

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  1. Introduction To Binomial Theorem A Binomial Expression Any algebraic expression consisting of only two terms is known as a binomial expression. It's expansion in power of x is shown as the binomial expansion. For example: (i) a + x Binomial Theorem (iii) 4x - 6y Such formula by which any power of a binomial expression can be expanded in the form of a series is known as binomial theorem. It can be easily understood by examples. 2 = a2 + 2ax + x2 2 Here, we see that the expression of (a + is simple, we just multiply (a + x) by (a + x). Expansion of (a + is little tougher, but what happens when the expansion is raised to the power of ten or more? So, we have to establish the formula for (a + x r, where n is any integer. Let us define 'a' as the first term, 'x' as the second term and 'n' as the exponent. The total terms in the expansion of (a + and (a + are 3 and 4 respectively, which means that the number of terms in the expansion is one more than the exponent. So total number of terms in the expansion (a + is (n + 1). Now, for n = 2 = (F.T.)n (S.T.)O + n/l! (S.T.)I + (S.T.)2 = a2 + 2ax + x2 Note: F. T. refers to first term i.e. Similarly, 3 3ax2 'a' and S.T. refers to second term i.e. 'x'
  2. When n is a positive integer, then Where n CO • n CI . n C2 • . are called Binomial coefficients. Properties of Binomial Expansion 1. There are (n + 1) terms in the expansion of (a + b)n, the first and the last term being an and bn respectively. If = ncy, then either x = y or x + y = n = > ncr = ncn = n!/r!(n-r)! 2. The general term in the expansion of (a + is (r + l)th term given as Tr+l =nCran-r + xr. Similarly the general term in the expansion of (x + is given as Tr+l = ncr xn-r ar. The terms are considered from the beginning. 3. The binomial coefficient in the expansion of (a + which are equidistant from the beginning and the end are equal i.e. ncr = ncn-r. — n +1Cr this concept will be discussed later in this Note: Here we are using ncr + ncr-l chapter. Also, we have replace by m+1Co because numerical value of both is same i.e. 1. Similarly we replace mcm by m+1Cm+1. Illustration: Find the expansion of (a-x)n. Solution: We know that = an + (n(n-1)/2!) a(n-2) +...+ xn. putting x = -x in the above expansion, we get, an + na(n-l) (-X) + (n(n-1)/2!) a(n-2) - na(n-l) X + (n(n-1)/2!) a(n-2) - Illustration: Find the value of (a+V(a2 -1))7 Solution:
  3. Here, we have to find the sum of two expansions whose terms are numerically the same, but in the second expansion the second, fourth, sixth and eight terms are negative, and therefore cancel the corresponding terms of the first expansion. Hence, the given expression = 2 {a 7 + 21 a 5 (a 2 - 1) + 35 a 3 (a 2 - 1) 2 + 7 a (a 2 _ 1)3} = 2a (64 a6- 112 + 56 - 7) Binomial Coefficients We know that, Let us write equation (i) in particular form by putting a = 1. Now it becomes, nco + nC1 X + nC2 + ...+ ncr xr + ...+ ncn xn Coefficients attached with different powers of x are called Binomial Coefficients. Now, again put a = 1 in the expression x + (n(n-1))/2! an-2 x2 + (n(n-1)(n-2))/3! an-3 x3 ....(n-r+l))/r! an-r xr + ... ...+ xn we get, If we compare coefficient of xr in expansions (ii) and (iii), we have, ncr = = n!/r!(n-r)! What does it mean when we compare two expansions? Since expansion (ii) is valid for any value of x, we can replace x by l/x , (x * 0) in it, we get,
  4. n CO + 'ICI l/ X + nC2 + Multiplying both sides by xn, it becomes, CO + xn-l + xn-r + ...+ ncr l/xr + Compare expansion (ii) and (iv). It will give us, that ncn-r, this is a beautiful result, which says that in the expansion of the equations of the like of (ii) the rth coefficient from the beginning is equal to the rthcoefficient from the end. Sum of Binomial Coefficients Putting x = 1 in the expansion (1+x)n = nco + nC1 X + nC2 + ...+ ncn. Co + nC1 X + nC2 + ncx X n, we get, We kept x = 1, and got the desired result i.e. inr=o Cr = 211. Note: This one is very simple illustration of how we put some value of x and get the solution of the problem. It is very important how judiciously you exploit this property of binomial expansion. Illustration: Find the Vallele Of co + C 2 + C 4 + Solution: We have, in the expansion of (l+x)n. nco - nCl X + nc2 - Now put x = -x; (I-xy = Now, adding both expansions, we get, ...+ (-1)n ncn xn. nc2 + nc4 + put = > (2+0)/2 = co + C2 +
  5. — 211-1 We have found the sum of binomial coefficients. But if these coefficients are multiplied by some factors can we find the sum for such expressions? Yes, we can often find it by creatively applying what we have learnt. Let us see how differentiation and integration are useful in these situations. Suppose we want to calculate the value of Cr i.e. O.CO + 1 Cl + 2C2 +...+ n ncn If we take a close look to the sum to be found, we find that coefficients are multiplied with respective powers of x. Coefficient Of A Particular Term In the expression of (a + x)n, the coefficient of second term is nCI, of the third term is nC2, of he fourth term is nC3 and so on. The suffix in each term being one less than the number of the term to which it applies; hence ncr is the coefficient of the (r+l)th term. This is called the General Term, because by giving different numerical values to r any of the coefficients may be found. Also, the indices of a and x in the (r + l)th term are expressable in terms of Thus, General term of the expansion (a + is dented as Tr+l = ncr an-r xr = Note 'General term is useful in ....(n-r+l))/r!) an-r xr. finding the following Particular Middle independent in (a) (b) Term Binomial of expression. term. term x. (d) Term containing the greatest coefficient of x. Illustration: Find the fifth term in the expansion of (a + Solution: General term of the expansion (a + x)'l is
  6. Tr+l = ncr an-r W. Now, for the given expression, required term is T5 = 17 C 4 a 13 = 38080 Illustration: Find the term independent of x in (1 + x + 2x3) ((3/2)x2 Solution: We want to find the term independent of x in product of (1 + x + and ((3/2)x2- (1/3x))9. So, we should find the terms containing xo, x-i and x-3 in the expansion (1/3x))9 and multiply them respectively with the appropriate term from (1 + x + i.e. constant term i.e. 1, term containing x i.e. 1 and the term containing i.e. 2. Now, Tr+l = ncr an-r xr [this is general term in (a+x)'l] 9cr (-(1/3x)Y This is general term in [ ((3/2)x2 - (1/3x))9] (i) To find the term containing xo we use 18 - 3r = 0, i.e. r = 6 so coefficient of xo we use (-1)6 9C6 3-3 2-3 = 9C6 (1/22.33). = 19/3 (E I) which is impossible. (ii) To find the term containing x-i we use 18-3r = (iii) To find the term containing x-3 we put 18-3r — 7. So, coefficient of x-3 in - = (-1)7 3-5 2-2 — -9C7 (1/23.35) . Therefore, required coefficient = 9C6 (1/23.33) x 1-2 x 9C7 (1/23.35) - 17/54. Middle Term (i) When n is even Middle term of the expansion is the (n/ 2 + l)th term i.e. ncn/2 an/2 in the expansion of (a + b)n. (ii) When n is odd
  7. Middle terms of the expansion are the (n/ 2 + l)th term and the ((n+3)/2)th term. These are given and in the expansion of (a + b)n. e.g. middle term in the expansion of (1+x)4 and (1+x)5. Expansion of (1+x)4 have 5 terms, so third term is the middle term which is the ((4/2)+ l)th term. Expansion of (1+x)5 have 6 terms, so 3rd and 4th both are the middle terms, which are the ((5+ and terms. Note: rth term from the end r + 2)th term from the beginning. (n * If there are two middle terms, then the binomial co-efficients of two middle terms will be equal and those two co-efficients will be greatest. Illustration: Find the middle term in the expansion of (1 Solution: - 2X + x2)n. We have (1 - 2x + x2)n = - (1 -x)2n. Here 2n is an even integer =>((2n/2) + l)th i.e. (n+l)th term will be the middle term. Now (-x)n. (n+ l)th term in (I-x)2n = 2ncn (1)2n-n (2n)!/n!n! Greatest Binomial Coefficient To determine the greatest coefficient in the binomial expansion, (l+x)n, when n is a positive integer. Coefficient of (Tr+1/Tr) = cr/Cr 1 Now the (r+l)th binomial coefficient will be greater than the rth binomial coefficient when, Tr+l > Tr But r must be an integer, and therefore when n is even, the greatest binomial coefficient is given by the greatest value of r, consistent with (1) i.e., r n/ 2 and hence the greatest binomial coefficient is ncn/2.
  8. To determine the positive integer. Consider Similarly in n be odd, the greatest binomial coefficient is given when, = (n-1)/2 or (n+1)/2 and the coefficient itself will be or both being are equal Note: The greatest binomial coefficient is the binomial coefficient of the middle term. Illustration: Show that the greatest the coefficient in the expansion of (x + 1/x)2n is (1.3.5...(2n Solution: Since middle term has the greatest coefficient. So, greatest coefficient = coefficient of middle term - 2ncn = = Numerically greatest term n-r+l . E = n can—r+lrv —1 numerically greatest term in the expansion of (a + where n is a Tr+l Thus ITr+11>lTrl If rca x x 1 x x 12 + I > a x r. Thus Tr+l will be the greatest term if, r Note : { ( (n +1)/ r) - 1} must be positive since n has the greatest value as per the equation (1). Illustration: Find the greatest term in the expansion of (3-2x)9 when x = 1. Some Important Results 1. Differentiating (1+x)n = Co + CIX + c2x2 + ...+ Cnxn of both sides we have,
  9. a constant) For x = 0, we get C — Therefore ((1+x)n+1 n-l = Cl + + + ...+ ncnxn-l. Put x = 1 in (E) so that (1211+1 = Cl + 2C2 + 3C3 + ...+ nCn. = -1 in (E) SO that O = Cl - 2C2 +...+ (-1)n-1 ncn Put x Differentiating (E) again and again we will have different results. 2. Integrating (1 + x Y, we have, + C = Cox + C + + (because ncr =n/r .n-1Cr-1) -1 r-_åc n-l (where C is = COX + Put x = (2+1 _ Put x Put x = cn. 1 in (F) and get n co + C 1/2 + ...Cn/n-kl . = -1 in (F) and get, 1/11+1 = CO - Cl/2 + C2/3 — 2 CO + 22/2 Cl + 22/3 C2 + ...+ 2 n +1/ n +1 2 in (F) and get, (31+1-1)/n+1 Problems Related to Series of Binomial Coefficients in Which Each Term is a Product of an Integer and a Binomial Coefficient, i.e. In the Form k.nCr. Illustration: If (1+x)n rxr then prove that Cl + 2C2 + 3C3 + . Solution: Method (i) : rth term of the given series rth term of the given series, tr = nCr r x n/r x n-1Cr-1 = n x n-1Cr-1 = > tr = + nCn = n 211-1 11-1 tr Sum of the series = Put x = 1 in the expansion of (1 n n so that
  10. (n-1C0 + n-1C1 + ...+ n-1Cn-1) = 2n-l nr-_Ã¥c = n.2n-1. Method (ii) : By Calculus We have (I + x) n = co + C 1 x + c2x2 + ...+ Cnxn. Differentiating (1) w.r.t. x, we get n-l = Cl + + 3C3 + Putting x = 1 in (2), we have, n 211-1 Illustration: + n Cnxn-l = Cl + 2C2 + + nCn. If (1+ X) n = xr then prove that Co + 2.C1 + 3.C2+... Application Expression 1. Divisibility problems: Of Binomial 1 + + + ...+ ncnxn. Let (1 + x) n = In any divisibility problem, we have to identify x and n. The number by which division is to be made can be, x, or x3, but the number in the base is always expressed in form of Illustration: Find the remainder when 7103 is divided by 24. Solution: 7103 = 7(50 _ 1)51 = 7(5051 _ 51C1 5050 + 51C2 5049 _ = 7 (5051 _ 51C15050 + ...+ 51C5050) -7 - 18 + 18 = 7(5051 - 51C15050 +...+ 51C5050) - 25 + 18 1)
  11. = > remainder is 18. 2. Questions involving the greatest integer function. The questions generally involves working with binomial expression on surds. Illustration: If I is the integral part and f is the fraction part of (2 + V 3 then prove that (l + f)(l- f) 1. Also prove that I is an odd integer. Solution: (2 + v 3)n, I + f where I is an integer and 0, f < 1. We have to show that I is odd and that (l + f)(l - f) Here note that (2 + v (2 + v =(4 - 3)'1 = 1 + V + It is thus required to prove that (2 + v But, (2 + v/ 3 + (2 + v/3)n = [2n - Cl. 2-1.&/3 + + [2n - Cl.2n-1.V3 + - C2.2n-2.3 + can-4.32 - Now O < (2 - V3) < 1 If (2 - V3)n = f' then I + f + f' _ even integer — Even Now ) < f < 1 and 0 < f' < 1 f + f' = integer (1) and (2) imply that f + f' I is odd and f' — -I-f Binomial expression with non- positive exponent The characteristics discussed hitherto were confined to positive integer n. other value, then binomial theorem is written as: If n takes any
  12. Say, we have to find out (x+a)n n 1+ If a > 1 then it is written as an (a + This is expanded as an + (x/ a (X/a)3 + an = .00] Since n is not positive integer therefore the series on the right hand side will converge only for Ix/al < 1. Moreover, there are infinite terms in the expansion contrary to the binomial expansion for a positive integer n. Solved Examples Example 1: Find the coefficient of the independent term of x in expansion of (3x - If the coefficient of (2r + 4)th and (r - 2)th terms in the expansion of (1+x) ncr = Solution: The general term of (3x independent of x if, 15-r-2r=0=>r Example 2: then find the value of r. Solution: is written, as Tr+l — 15Cr (3x)15-r (-2/x2)r. It is 310 25 8are equal The general term of (1 + is Tr+l = Crxr Hence coefficient of (2r + 4)th term will be — 18C2r+3 T2r+4 = T2r+3+l and coefficient or (r - 2)th term will be — 18Cr-3. — > 18C2r+3 — 18Cr-3. = > (2r + 3) + (r-3) - 18
  13. Example 3: If al, a2, a3 and a4 are the coefficients of any four consecutive terms in the expansion of (1+x)n then prove that: al/ (al-pa2) + a2/(a3+a4) = 2a2/(a2-å-a3) Solution: As al, a2, a3 and a4 are coefficients of consecutive terms, then Let al — a3 ncr ncr+2 and a4 = ncr+3 Now al/ (al+a2) — ncr/(ncr+ncr+l) Similarly, a2/(a2+a3) NOW a3/(a34-a4) + a 1/ (a 1-l-a2) = 2 (r +1)/ (n +1) = 2a2/(a2-pa3)

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