IIT JEE , AIEEE Forum - Mathematics , Algebra

sahilgupta_iit

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

vasanth

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)C_r_-1 = 17C₂ = 136

vasanth

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

pink_ele

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

pink_ele

arey yaar............its simply.........initially divding n multiplying by
2n+2)(2n+4).................4n
den div n mul by...........n!.......

rudra.panda

any answers to the first one?

hpudipeddi

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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