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  1. #1
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    Hi >> I want your help

    Hi all

    I want check my anwsers


    3 If r = -1 and frist term is 1/2 find the sum of the 10 terms of a Gp

    4 - Use mathematical induction to prove that 2 + 4 + 6 ..+ 2n = n(n+1)

    5 Expand ( 2x 3y)^4 by using binomial formula

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  2. #2
    Junior Member SuperCalculus's Avatar
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    Quote Originally Posted by r-soy View Post
    Hi all

    I want check my anwsers


    3 – If r = -1 and frist term is 1/2 find the sum of the 10 terms of a Gp

    4 - Use mathematical induction to prove that 2 + 4 + 6 ……..+ 2n = n(n+1)

    5 – Expand ( 2x – 3y)^4 by using binomial formula

    I can only really help with the last one, but as far as that's concerned this is how I do it;

    4th bimomial row; 1, 4, 6, 4, 1

    So,
    (1)(-3y)^4 + (4)(-3y)^3(2x) + (6)(-3y)^2(2x)^2 + (4)(-3y)(2x)^3 + (1)(2x)^4

    = 81y^4 - 216xy^3 + 216x^2y^2 - 216x^3y + 16x^4 if my multiplication is correct.
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  3. #3
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    Hello, r-soy!

    3) If r = -1 and first term is \tfrac{1}{2},
    find the sum of the first 10 terms of the GP.
    \text{Sum of first }n\text{ terms: }\;S_n \;=\;a\,\frac{1-r^n}{1-r}


    We have: . a = \tfrac{1}{2},\;r = -1,\;n=10

    Therefore: . S_{10} \;=\;\frac{1}{2}\cdot\frac{1-(-1)^{10}}{1-(-1)} \;=\;\frac{1}{2}\cdot\frac{1-1}{2} \;=\;0




    4) Use mathematical induction to prove that: . 2 + 4 + 6 + \hdots + 2n \:=\: n(n+1)
    Verify S(1)\!:\;\;2 \:=\:1(1+1) . . . True.


    Assume S(k)\!:\;\;2 + 4 + 6 + \hdots + 2k \:=\:k(k+1)


    Add 2(k+1) to both sides: . 2 + 4 + 6 + \hdots + 2(k+1) \;=\;k(k+1) + 2(k+1)


    And we have: . 2 + 4 + 6 + \hdots + 2(k+1) \;=\;(k+1)(k+2)


    We have proved S(k+1) . . . The proof is complete.




    5) Expand: . (2x - 3y)^4
    We have: . (2x-3y)^4

    . . . =\; {4\choose4}(2x)^4(-3y)^0 + {4\choose3}(2x)^3(-3y)^1 + {4\choose2}(2x)^2(-3y)^2  + {4\choose1}(2x)^1(-3y)^3 + {4\choose0}(2x)^0(-3y)^4

    . . . =\quad\; 1\cdot16x^4\cdot1 \quad\;\; +\quad 4\cdot8x^3(-3y) \quad+\quad\;\; 6\cdot4x^2\cdot 9y^2 \quad\; + . 4\cdot2x(-27y^3) \;\;+\;\;\; 1\cdot1\cdot81y^4

    . . . =\quad\; 16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4

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  4. #4
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    For (1) you have used r= 10 in the denominator when r= -1.

    (And, by the way, if it were 10, 1- 10= -9, not -11!)

    For (2) you are fine until you get to k^2+ 3k+ 2 and there you stop. Don't forget what you are trying to prove. You want to prove that sum is n(n+1) which, for n= k+1, is (k+1)(k+2). Factor k^2+ 3k+ 2!

    For (3), (2x+3y)^4 has mysteriously become (3x+ 3y)^4!
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