2305Answer 5:
Number of terms in (x1+x2+x3.....+xk)n is -n+r-1Cr-1
Answer - n+3C3
ATTENTION Q7 ) added
i'm weak in algebra Pleasse help.So Posting the questionsVERY VERY WEAK
Q1) Prove that 22225555 + 55552222 is divisible by 7
Q2) If 7 divides 323232 then find the remainder . ans is 4
Q3) Show that 19921998-19551998-19381998+19011998 is divisible by 1998.
Q4) If (1-x3)n = \sum_{r=o}^{n}{a_{r}x^{r}(1-x)^{3n-2r}}, then find ar,where n\epsilon N ^{n}C_{r}3^{r}
Q5) find the number of terms in the expansion of ( x + y + z + w )n , n\epsilon N ans ^{n+3}C_{3}
Q6) Show that [(√3+1)2n]+1 is divisible by 2n+1 for all n belongs to Natural no. , where [.] is greatest integer function.
Q7) If a,b,c,d be four consecutive binomial coefficients in the expansion of (1 + x)2 then prove that (bb+c)2>ac(a+b)(c+d) if x > 0
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26 Answers
21\\\textup{Q3) } \textup{let } N=1992^{1998} - 1955^{1998} - 1938^{1998} + 1901^{1998}\\ \textup{we know } a^n-b^n \textup{ is always divisible by (a-b) for any +ve integer n}\\ \therefore \; 1992^n - 1938^n \textup{ is divisible by (1992-1938)i.e 54.}\\ \textup{similarly } 1955^n- 1901^n \textup{ is divisible by (1955-1901)=54} \\\therefore \textup{ N is divisible by 54}\\ \textup{Now Similarly } 1992^n - 1955^n \textup{ is divisible by (1992-1955)= 37}\\ \textup{ and}\\ 1938^n - 1901^n \textup{ is divisible by (1938-1901)= 37 }\\ \textup{therefore N is divisible by 37}\\ \textup{Now since N is divisible by 54 and 37 it has to be divisible by LCM(54,37)=54.37=1998}
106to the last question .... zero (provided a,b,c....,h,k, A,B,C,....,H are real)
23052305Answer 2:
32 ≡ 2 (mod 3)
3232 ≡ (26)5*4 ≡ 645*4 ≡ 4≡1(mod 3)
Let 3232 = 3k + 1
323232 ≡ (323)k*32 ≡ 64k * 32 ≡ 32 ≡ 4(mod 7)
Answer - 4
2305Answer 1:
1111 ≡ 5 (mod 7)
2222 ≡ 10 (mod 7)
22225555 ≡ 1000001111 (mod 7)
5555 ≡ 25 (mod 7)
55552222 ≡ 6251111 (mod 7)
Therefore, 22225555 + 55552222≡(100000+625)≡0( mod 7)
21ok for 2nd here is teh link-
http://www.targetiit.com/iit-jee-forum/posts/remainder-11788.html
1sorry asish there was a typo got it edited.
SORRY AGAIN [3][3]
106Q6. i think there is a typo
Q2. something missing at end of question
1z= ( a - i b ) ^3
x + i y = ( a - i b ) ^3
on cubing r.h.s and equating real and imaginary terms we get
x = a^3 - 3a b^2
y = b^3 - 3a^2 b
\frac{x}{a} = a^2 - 3b^2
\frac{y}{b} = b^2 - 3a^2
\frac{x}{a} -\frac{y}{b} = a^2 - 3b^2 - ( b^2 -3a^2)
{\color{red} \frac{x}{a} - \frac{y}{b}= 4( a^2 - b^2 )}
39Q. If alpha, beta, gamma are cube roots of p where p < 0, for any x, y, z find the value of
\frac{x\alpha +y\beta + z\gamma }{x\beta +y\gamma +z\alpha}
39On request of gr8dreams, I'm posting some more conceptual complex number problems.
Q. If aeix, beiy, ceiz are three collinear points, find a trigonometric relation which satisfies x,y,z.
Q. If we are given that
z = x + iy
z1/3 = a - ib
Prove that
\frac{x}{a} - \frac{y}{b} = 4(a^2 - b^2)
21well these probs hav been hashed, re hashed, re-re hashed many a times here...
(i heard hsbhatt sir saying this to someone on goiit :P)
Q1) http://targetiit.com/iit-jee-forum/posts/reasoning-type-9087.html
Q2) this has also been done....though not finding the link rit now
39Here's an easy one....
Q. Find the number of imaginary roots in the equation -:
\frac {A^2}{x-a} + \frac{B^2}{x-b} + \frac{C^2}{x-c} +....+\frac{H^2}{x-h} = k
106no. of terms in (x+y+z+w)n
=> Let power of x be a, y be b, z be c and w be d
Then acc. to multinomial theorem,
a+b+c+d =n n≥a,b,c,d≥0
=> coeff of xn in (1+x+x2+...)4
= (1-x)-4
= n+4-1Cn which is the answer
39No ut10...your approach is right but answer is not...try again :)
39Try this one...an addition to your algebra thread lol
Q. If
a = 1 + \frac{x^3}{3!} + \frac{x^6}{6!} + ......\infty
b = x + \frac{x^4}{4!} + \frac{x^7}{7!} + ......\infty
c = \frac{x^2}{2!} + \frac{x^5}{5!} + \frac{x^8}{8!} + ......\infty
Find the value of
a^3 + b^3 + c^3 - 3abc