{"id":3483,"date":"2023-10-05T10:41:46","date_gmt":"2023-10-05T10:41:46","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3483"},"modified":"2023-10-05T10:42:53","modified_gmt":"2023-10-05T10:42:53","slug":"week-3-ss2-third-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-3-ss2-third-term-further-mathematics-notes\/","title":{"rendered":"Week 3 – SS2 Third Term Further Mathematics Notes"},"content":{"rendered":"

WEEK THREE
\n<\/strong>TOPIC:BINOMIAL EXPANSION: PASCAL TRIANGLE, BINOMIAL THEOREM OF NEGATIVE, POSITVE AND FRACTIONAL POWER
\n<\/strong>PASCAL’S TRIANGLE
\n<\/strong>Consider the expressions of each of the following:
\n(x + y)0<\/sup>; (x + y )1<\/sup>; (x + y)2<\/sup>; (x + y)3<\/sup>; (x + y)4
\n<\/sup>(x + y)0<\/sup> = 1
\n(x + y)1<\/sup> = 1x + 1y
\n(x + y)2<\/sup> = 1x2<\/sup> + 2xy + 1y2<\/sup>
\n\t\t(x + y)3<\/sup>= 1x3<\/sup> + 3x2<\/sup>y + 3xy2<\/sup> + 1y3<\/sup>
\n\t\t(x + y)4<\/sup> = 1x4<\/sup> + 4x3<\/sup>y + 6x2<\/sup>y2<\/sup> + 4xy3<\/sup> + 1x4
\n<\/sup>The coefficient of x and y can be displayed in an array as:
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01
\n\u00a0\u00a0\u00a0\u00a01\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01
\n1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a06\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01
\nThe array of coefficients displayed above is called Pascal’s triangle, and it is used in determining the co-efficients of the terms of the powers of a binomial expression
\nCoefficient of (x + y)0<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01
\nCoefficient of (x + y)1<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01
\nCoefficients of (x + y)2<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01
\nCoefficients of (x + y)3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01
\nCoefficients of (x + y)4<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a06\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01<\/p>\n

\u00a0Example 1
\n<\/strong>Using Pascal’s riangle, expand and simplify completely: (2x + 3y)4<\/sup>
\n\t\tSolution:
\n<\/strong>(2x + 3y)4<\/sup> = (2x)4<\/sup> + 4(2x)3<\/sup> (3y) + 6(2x)2<\/sup>(3y)2<\/sup> + 4(2x)(3y)3<\/sup> + (3y)4<\/sup>
\n\t\t= 16x4<\/sup> + 96x3<\/sup>y + 216x2<\/sup>y2<\/sup> + 216xy3<\/sup> + 81y4<\/sup><\/p>\n

\u00a0Examples 2:
\n<\/strong>Using pascal’s triangle, the coefficients of (x + y)5<\/sup>are: 1,5,10,10,5,1.
\nTherefore (x \u2013 2y)5<\/sup>\u00a0\u00a0\u00a0\u00a0= x5<\/sup> + 5x4<\/sup>(-2y) + 10x3<\/sup>(-2y)2<\/sup> + 10x2<\/sup>(-2y)3<\/sup> + 5x(-2y)4<\/sup> + (-2y)5
\n<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sup>= x5<\/sup> \u2013 10x4<\/sup>y + 40x3<\/sup>y2<\/sup> \u2013 80x2<\/sup>y3<\/sup> + 80xy4<\/sup> \u2013 32y5<\/sup>
\n\t\tExample 3
\n<\/strong>Using Pascal’s triangle, simplify, correct to 5 decimal places (1.01)4
\nSolution
\n<\/strong>We can write (1.01)4<\/sup> = (1 + 0.01)4<\/sup>
\n\t\t(1 + 0.01)4<\/sup> = 1 + 4(0.01) + 6(0.01)2<\/sup> + 4(0.01)3<\/sup>+(0.01)4 <\/sup>
\n\t\t= 1 + 0.04 + 0.0006 + 0.000004 + 0.00000001
\n= 1.04060401
\n= 1.04060 (5 d.p)
\nThe Binomial Expansion Formula
\nConsider the expansion of (x + y)5<\/sup> again
\n(x + y)5<\/sup> = (x + y)(x + y)(x + y)(x + y)(x + y)
\nThe first term is obtained by multiplying the xs in the five brackets. there is only one way to doing this
\n(x + y)n<\/sup> = xn + nxn \u2013 1<\/sup>y + xn<\/sup> \u2013 2<\/sup>y2<\/sup> + xn-3<\/sup>y3<\/sup> + \u2026.
\nxn-r<\/sup>yr<\/sup> + \u2026. yn<\/sup>
\n\t\tIt can be shown that the binomial expansion formula holds for positive, negative, integral or any rational value of n, provided there is a restriction on the values of x and y in the expansion of (x + y)n<\/sup>
\n\t\t\t<\/sub>We shall however consider only the binomial expansion formula for a positive integral n<\/p>\n

\u00a0Example 4:
\n<\/strong><\/p>\n

    \n
  1. Write down the binomial expansion of 6<\/sup> simplifying all the terms\n<\/li>\n
  2. \n
    Use the expansion in (a) to evaluate (1.0025)6<\/sup> correct to five significant figures.\n<\/div>\n

    Solution
    \n<\/strong>6<\/sup> = 1 + 6<\/sup>C1 <\/sub>1<\/sup> + 6<\/sup>C2<\/sub>2
    \n<\/sup> + 6<\/sup>C3 <\/sub>3 <\/sup> + 6<\/sup>C4 <\/sub>
    \n\t\t\t\t\t<\/sup> + 6<\/sup>C5 <\/sub>5 <\/sup> + 6<\/sup>C6 <\/sub>6
    \n<\/sup>6<\/sup>= 1 + x + 3<\/sup> + x
    \n4<\/sup> + x 5 <\/sup> + 6
    \n<\/sup>= 1 + x + x2<\/sup>+ x3<\/sup> +x4<\/sup> + x5<\/sup> + x6<\/sup>\n\t\t\t\t<\/li>\n

  3. \n
    (1.0025)6<\/sup> = (1 + 0.0025)6<\/sup>\n\t\t\t\t<\/div>\n

    = )6<\/sup>
    \n\t\t\t\t= )6<\/sup>
    \n\t\t\t\tPut x =
    \nx = x 4 = = 0.01
    \ntherefore (1.0025)6<\/sup> = 1 + (0.01) + (0.01)2<\/sup> + (0.01) +(0.01)4<\/sup> + \u2026
    \n= 1 + 0.015 + 0.00009375 + 0.0000003125
    \n= 1.0150940625
    \n= 1.0151 (5 s.f.)<\/p>\n

    \u00a0EVALUATION
    \n<\/strong>Expand ( 2 + 4x )4<\/sup> simplifying the terms<\/p>\n

    \u00a0Example 5<\/strong>(a) Using the binomial theorem, obtain the expansion of (1 + 3x)6<\/sup> + (1 \u2013 3x)6<\/sup> simplifying all the terms
    \n(b)Use the above result to calculate the value of (1.03)6<\/sup> + (0.97)6<\/sup>, correct to five decimal places
    \nSolution:
    \n(1 + 3x)6<\/sup> = 1 + 6<\/sup>C1<\/sub> (3x) + 6<\/sup>C2<\/sub> (3x)2<\/sup> + 6<\/sup>C3<\/sub> (3x)3<\/sup> _ 6<\/sup>C4<\/sub> (3x)4<\/sup> _ 6<\/sup>C5<\/sub> (3x)5<\/sup> + 6<\/sup>C6<\/sub> (3x)6<\/sup> \u2026.. (1)
    \n(1 – 3x)6<\/sup> = 1 – 6<\/sup>C1<\/sub> (3x) + 6<\/sup>C2<\/sub> (3x)2<\/sup> – 6<\/sup>C3<\/sub> (3x)3<\/sup> + 6<\/sup>C4<\/sub> (3x)4<\/sup> _ 6<\/sup>C5<\/sub> (3x)5<\/sup> + 6<\/sup>C6<\/sub> (3x)6<\/sup> \u2026.. (2)
    \nAdding (1) and (2)
    \n(1 + 3x)6<\/sup> +(1 – 3x)6<\/sup> = 2 + 2 x 6<\/sup>C2<\/sub> (3x)2<\/sup> + 2 x 6<\/sup>C4<\/sub> (3x)4<\/sup> + 2 x 6<\/sup>C6<\/sub> (3x)6<\/sup>
    \n\t\t\t\t= 2 + 2 x 9x2<\/sup> + 2 x x 81x4<\/sup> + 2 x 729x6<\/sup>
    \n\t\t\t\t= 2 + 270x2<\/sup> + 2430x4<\/sup> + 1458x6<\/sup>
    \n\t\t\t\t(1.03)6<\/sup> = (1 + 0.03)6<\/sup>
    \n\t\t\t\t(0.97)6<\/sup> = (1 \u2013 0.03)6<\/sup>
    \n\t\t\t\tPut 1 + 0.03 = 1 + 3x
    \nTherefore 3x = 0.03
    \nTherefore x = 0.01
    \nHence
    \n(1.03)6<\/sup> + (0.97)6<\/sup> = 2 + 270(0.01)2<\/sup> + 2430(0.01)4<\/sup> + 1458(0.01)
    \n= 2 + 0.027 + 0.0000243 + 2.0270243
    \n= 2.02702 (5 d.p)<\/p>\n

    \u00a0Example 6
    \n<\/strong><\/li>\n

  4. Using the binomial theorem, expamd (1 + 2x)5<\/sup>, simplifying all the terms\n<\/li>\n
  5. \n
    Use your expansion to calculate the value of 1.025<\/sup>, correct to six significant figures\n<\/div>\n

    If the first three terms of the expansion of (1 + px)n<\/sup> in ascending powers of x are 1 + 20x + 160x,
    \nFind the values of n and p
    \nSolution:\n<\/li>\n

  6. \n
    (1 + 2x)5<\/sup> = 1 . 5<\/sup>C1<\/sub>(2x) + 5<\/sup>C2<\/sub>(2x)2<\/sup> + 5<\/sup>C3<\/sub>(2x)2<\/sup> + 5<\/sup>C4<\/sub>(2x)4<\/sup> + 5<\/sup>C5<\/sub>(2x)5<\/sup>\n\t\t\t\t<\/div>\n

    = 1 + 5.(2x) + . 4x2<\/sup> + . 8x3<\/sup> + . 16x + 32x5<\/sup>
    \n\t\t\t\t= 1 + 10x + 40x2<\/sup> + 80x3<\/sup> + 80x4<\/sup> + 32x5<\/sup>\n\t\t\t\t<\/li>\n

  7. \n
    (1.02) = (1 + 0.02)\n<\/div>\n

    Put 1 + 0.02 = 1 + 2x
    \nTherefore 2x = 0.02
    \nx = 0.01
    \nHence:
    \n(1.02)5<\/sup> = 1 + 10(0.01) + 40(0.01)2<\/sup> + 80(0.01)3<\/sup> + 80(0.01)4<\/sup> + 32(0.01)5<\/sup>
    \n\t\t\t\t= 1 + 0.1 + 0.004 + 0.0008 + 0.00000008
    \n= 1.10408 (6.s.f.)
    \n6.3 The Binomial Theorem for any index
    \nThe Binomial expansion formula is also applicable to any index n, where n can be positive or negative integer or even a fraction
    \nIf \/x\/ 1, then:
    \n(1 + x)n<\/sup> = 1 + nx + + + x4<\/sup> + \u2026 where n may be a negative integer or a fraction.<\/p>\n

    \u00a0Example 7
    \n<\/strong>Use the Binomial expansion formula to obtain the first five terms of the expansion of (1 + x)-2<\/sup>
    \n\t\t\t\tSolution:
    \n<\/strong>(1 + x)-2<\/sup> = 1 + (-2) ( x) + ( ( x)2<\/sup> + ( x)3<\/sup> + ( x)4<\/sup> + \u2026.
    \n= 1 \u2013 x + 3. 2<\/sup> – 4.3<\/sup> + 5.4<\/sup><\/p>\n

    \u00a0<\/li>\n

  8. \n
    (1 + px)n<\/sup> = 1 + 20x + 160x2<\/sup> + \u2026\n<\/div>\n

    (1 + px)n<\/sup> = 1 + n<\/sup>c1<\/sub> (px) + n<\/sup>c1<\/sub> (px)2<\/sup> + \u2026
    \n = 1 + npx + p2<\/sup>x2<\/sup> \u2026
    \n = 1 + 20x + 160x2<\/sup> + \u2026
    \nBy equating coefficients
    \nnp = 20 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2026 \u00a0\u00a0\u00a0\u00a0(1)
    \np2<\/sup> = 160\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2026\u00a0\u00a0\u00a0\u00a0(2)
    \nFrom (1) p = \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2026\u00a0\u00a0\u00a0\u00a0(3)
    \nTherefore p2<\/sup> = \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2026\u00a0\u00a0\u00a0\u00a0(4)
    \nSubstituting (4) into (2)
    \n x = 160
    \nx 200 = 160
    \nThere 200(n \u2013 1) = 160n
    \n200n \u2013 200 = 160n
    \n200n \u2013 160n = 200
    \n 40n = 200
    \n n = 5
    \nFrom (3)p = = 4
    \nHence, n = 5, p = 4<\/p>\n

    \u00a0Example 8
    \n<\/strong>Obtain the first four terms of the explanation of (2 + x)8<\/sup>in ascending powers of x. hence, find the value of (2.005)8<\/sup>, correct to five significant figures.
    \nSolution:
    \n(2 + x)8<\/sup>= 28<\/sup>(1+ x)8
    \n<\/sup>= 28<\/sup>[1 +8<\/sup>C1<\/sub>( x) + 8<\/sup>C2 <\/sub>( )2<\/sup> + 8<\/sup>C3 <\/sub>( )3 <\/sup> + \u2026 ]
    \n= 28<\/sup>[1 +8( x) + ( )2<\/sup> + ( )3 <\/sup> + \u2026]
    \n= 28<\/sup>[1 + 2X + X2<\/sup>+ X3<\/sup> + \u2026]
    \nWrite 2.0.005
    \nPut 2 + x = 2 + 0.005
    \nTherefore x = 0.005
    \nTherfore x = 0.005 x 2
    \n = 0.01
    \nHence,
    \n(2.005)8<\/sup> = 28<\/sup>[1 +2(0.01) + (0.01)2<\/sup>+ (0.01)3 <\/sup>]
    \n(2.005)8<\/sup> = 28<\/sup> + 29<\/sup>(0.01) + 26<\/sup>.7(0.01)2<\/sup> + 25<\/sup> x 7(0.01)3<\/sup> + \u2026
    \n\t\t\t\t\t<\/sup>= 256 + 5.12 + 0.0448 + 0.000224
    \n = 261.165025
    \n = 261.17 (5 s.f.)<\/p>\n

    \u00a0GENERAL EVALUATION<\/strong>
    \n\t\t\t\t1) Write down and simplify all the terms of the binomial expansion of ( 1 \u2013 x )6<\/sup> . Use the expansion to evaluate 0.9976<\/sup> correct to 4 dp
    \n2) Write down the expansion of ( 1 + \u00bc x ) 5<\/sup> simplifying all its coefficients
    \n3) Use the binomial theorem to expand ( 2 \u2013 \u00bc x)5<\/sup> and simplify all the terms
    \n4) Deduce the expansion of ( 1 \u2013 x +x2<\/sup> )6<\/sup> in ascending powers of x<\/p>\n

    \u00a0Reading Assignment<\/strong>
    \n\t\t\t\tNew Further Maths Project 2 page 73 \u2013 78<\/p>\n

    \u00a0WEEKEND ASSIGNMENT
    \n<\/strong>If the first three terms of the expansion of ( 1 + px )n<\/sup> in ascending powers of x are 1 + 20v + 160x find the value of
    \n1) n a) 2 b) 3 c) 4 d) 5
    \n2) p a) 2 b) 3 c) 4 d) 5
    \n3) In the expansion of ( 2x + 3y )4<\/sup> what is the coefficient of y4<\/sup> a) 16 b) 81 c) 216 d) 96
    \n4) How many terms are in the expansion of ( 1 \u2013 4x ) 5<\/sup> a) 3 b) 5 c) 6 d) 8
    \n5) What is the third term in the expansion of ( 1 \u2013 3x )6<\/sup> in ascending powers of x a) 18 b) -540 c) 135 d) 729<\/p>\n

    \u00a0THEORY
    \n<\/strong>1) Using binomial theorem, write down and simplify the first seven terms of the expansion of ( 1 + 2x )10<\/sup> in ascending powers of x
    \n2) Expand ( 2 + x )5<\/sup> ( 1 \u2013 2x ) 6<\/sup> as far as the term in x3<\/sup> . Evaluate ( 1.999 )5<\/sup> ( 1.002 )6<\/sup><\/p>\n

    \u00a0
    \n\t\t\t\t\u00a0<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

    WEEK THREE TOPIC:BINOMIAL EXPANSION: PASCAL TRIANGLE, BINOMIAL THEOREM OF NEGATIVE, POSITVE AND FRACTIONAL POWER PASCAL’S…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,275],"tags":[],"class_list":["post-3483","post","type-post","status-publish","format-standard","hentry","category-posts","category-third-term-ss2-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3483","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3483"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3483\/revisions"}],"predecessor-version":[{"id":3484,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3483\/revisions\/3484"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3483"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3483"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3483"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}