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14 Nov 2011 11:03:26 IST
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S = 3+7x+13x^2+21x^3+31x^4+......(infinity) Find S if x = 1/2
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14 Nov 2011 13:40:34 IST
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S = 3 + 7/2 + 13/4 + 21/8 ...... S/2 = 3/2 + 7/4 + 13/8 +..... S/2 - 3 = 4/2 + 6/4 + 8/8 ..... on subtracting the latter from the former. this is an AGP. therefore S/2 - 3 = 4/(1-1/2) + [2(1/2)]/[1-1/2]^2 or, S/2 - 3 = 8 + 1/[1/4] = 12 so, S /2 = 15 S = 30. sum of AGP untill infinity = a/(1-r) + dr/(1-r)2 where a is the first term of the AP , r is the common ratio of GP and d is the common difference of AP,.
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mathematics is the food of my soul.
There are 3 types of people in the world. Those who can count and those who can't . :-)
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14 Nov 2011 17:43:06 IST
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The answer is 18..
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14 Nov 2011 18:11:59 IST
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S = 3 + 7x + 13x2+ 21x3 +......... (i) Sx = 3x + 7x2 + 13x3+........ (ii) (i) - (ii) S(1-x) =3 + 4x + 6x2 + 8x3........ (iii) S(1-x)x = 3x + 4x2 + 6x3 + 8x4........ (iv) (iii) - (iv) S(1-x)2= 3 + x + 2(x2+ x3+ x4+........) = 3 +x + 2[x2/(1-x)] putting x=1/2 S/4 =3 + 1/2 + 1= 9/2 S = 18 ans. __/\__
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14 Nov 2011 18:14:33 IST
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S = 3 + 7x + 13x2+ 21x3 +......... (i) Sx = 3x + 7x2 + 13x3+........ (ii) (i) - (ii) S(1-x) =3 + 4x + 6x2 + 8x3........ (iii) S(1-x)x = 3x + 4x2 + 6x3 + 8x4........ (iv) (iii) - (iv) S(1-x)2= 3 + x + 2(x2+ x3+ x4+........) = 3 +x + 2[x2/(1-x)] putting x=1/2 S/4 =3 + 1/2 + 1= 9/2 S = 18 ans. __/\__
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15 Nov 2011 11:50:45 IST
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@ shrey i think i have done it wrong. sorry my fault.. @ prahlad : good going buddy/////
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mathematics is the food of my soul.
There are 3 types of people in the world. Those who can count and those who can't . :-)
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15 Nov 2011 15:59:26 IST
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i got what i did wrong. S = 3 + 7/2 + 13/4 + 21/8 ...... S/2 = 3/2 + 7/4 + 13/8 +..... S/2 - 3 = 4/2 + 6/4 + 8/8 ..... on subtracting the latter from the former. this is an AGP. but we have to add 2/1 on both the sides to complete it. then use the formula i have mentioned...
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mathematics is the food of my soul.
There are 3 types of people in the world. Those who can count and those who can't . :-)
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1 Feb 2012 12:38:24 IST
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S = 3 + 7x + 13x2+ 21x3 +......... (A)
Multiplying both sides by 'x'
Sx = 3x + 7x2 + 13x3+........ (B) Subtracting (B) from (A), S(1-x) =3 + 4x + 6x2 + 8x3........ (C) again, multiplying both sides by 'x'
S(1-x)x = 3x + 4x2 + 6x3 + 8x4........ (D) Subtracting (D) from (C) S(1-x)2 = 3 + x + 2(x2+ x3+ x4+........) = 3 +x + 2[x2/(1-x)]
S = { 3 +x + 2[x2/(1-x)] } / { (1-x)2 } substituting the value, x=1/2 , in the above equation, we have
S = 18 |
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