3put x=1 and differentiate the expression and then put x=1 solves i think.
Paragraph Question
(1+px+x2)n = 1+a1x + a2x2 +.....+a2nx2n
The remainder when a1 + 5a2+ 9a3 + 13a4 + ......+(8n - 3)a2n is divided by p + 2 is
1. 1
2. 2
3. 3
4. 0
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8 Answers
62This one is a good question.. It took me some time to get hold of what was going on:
A hint for you...
ar=a2n-r
Now try This .. otherwise I will solve it :)
4
Diff the expansion & substituting x=1
sigma ]1 to 2n [
(4r-3)ar = (4n-3)(p+2)n + 3
So, remainder is 3
362\\(1+px+x^2)^n=1+a_1x+a_2x^2+...+a_nx^{2n} \\S=a_1+5a_2+9a_3+..+(8n-3)a_{2n} \\S=(8n-3)a_{1}+.....+(1)a_{2n} \\2S=(8n-2)[a_1+a_2+a_3+..+a_{2n}] \\S=(4n-1)[a_1+a_2+a_3+..+a_{2n}] = 4(1+p+1)^n \text{Puttin x=1 in the first equation} \\S=(4n-1)[a_1+a_2+a_3+..+a_{2n}] = 4(2+p)^n
So the remainder is zero :)
21this q if i m not wrong was asked in previous year's fiitjee's part test 3
23is the following substitution correct????
n=1, p =-1, x= 1
then 1= 1+a1
thus a1 = 0
and we will need to find the remainder when p+2 = 1 divides a1
wich is zero
4@Nishant Sir : Thanks
@Deepak: This qs was from a book.I don't remember abt fiitjee
@Qwerty : Ya By substitution we'll get the ans but i wanted the solution