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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jun 2008 15:29:03 IST
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Q.1. Find the sum 1! + 2! + 3! +4! +..................+ n!.
Q.2. The number of integral solutions of x + y + z = 0 with 
Q.3. Let x = (2n+1)(2n+3).........(4n-3)(4n-1) and y = (1/2)n (4n)! n! / (2n)!(2n)! then x - y + 2n is equal to
(a) 0 (b) (2n)!/2n (c)  (d) none of these
Q.4. The units digit of 172005 + 112005 - 72005 is...
...Rates assured
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22 Jun 2008 15:36:42 IST
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2)
let
a = x + 5
b= y +5 nd c = z+ 5
so we've
a + b + c =15............nd a,b,c >= 0
usin multinomial theorem......no.of integral solutions
(n + r - 1)Cr-1 = 17C2 = 136
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dbznfreak---watchin episodes for 6 yrs--movin on to dbgt
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22 Jun 2008 15:41:35 IST
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4)
171 -----7 (units digit) 175-----7
172 ----9
173 --- 3
174----1
so u get a pattern.....following this-----units digit in the 2005th power of 17 is-----7
ALL POWERS OF 11 END WITH ------- 1
7 follows 17's pattern---------so units digit is 7
answer is = 7 + 1 - 7 = 1
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dbznfreak---watchin episodes for 6 yrs--movin on to dbgt
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22 Jun 2008 15:45:07 IST
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q.3....if u see.................x=(2n+1)(2n+3).........(4n-3)(4n-1)
=(2n+1).....................4n/2^n(n+1).........2n)
=2n!(2n+1).....................4n/2^n(n+1).........2n)2n!
=4n!n!/2n!2n!2^n=y
hence none of dese
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nobody is wrong
even a stopped clock is right twice a day |
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22 Jun 2008 17:46:38 IST
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arey yaar............its simply.........initially divding n multiplying by
2n+2)(2n+4).................4n
den div n mul by...........n!.......
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nobody is wrong
even a stopped clock is right twice a day |
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22 Jun 2008 17:50:27 IST
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any answers to the first one?
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God does not care about our mathematical difficulties. He integrates empirically. ~~~Albert Einstein (1879-1955)~~~~
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.~~~Pierre de Fermat (1601-1665)~~~
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22 Jun 2008 18:54:08 IST
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4)If u observe carefully there is a repetition in the units digit the powers of each one:
17^1=7
17^2=9
17^3=3
17^4=1
Therefore in 17 there is a repetition in the units digit,for an incease of 4 in the power the units digit repeats:
Therefore units digit of 17^2005=2005/4 reminder is 1
the units digit is the units digit of 17^1 i.e 7
7 also follows that pattern therefore the units digit of 7^2005 is 7
The units digit of all the powers of 11 is 1
Therefore units digit is
7+1-7=1
Therefore units digit is 1
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