211)
Considering in pairs
(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6).............(99-100)(99+100)+101^{2}
now taking -1 common in each bracket....we get
-1(1+2+3+4.........100)+101^2
=-1(\frac{100.101}{2})+101^2
=5151
Q1 x_n=1^2-2^2+3^2-....+(-1)^{n-1}n^2.
Find x101
Q2 Find no of roots of
(2.4)x=(2.6)x-1
Graphically it appears that one soln...but i am loooking for algebraic soln...
Q3 If a,b,c ε R and (b-1)2<ac then find no. of solns of eqns
\begin{matrix} ax_1^2+2bx_1+c=2x_2\\ ax_2^2+2bx_2+c=2x_3\\ ax_3^2+2bx_3+c=2x_4\\ ...................\\ ...................\\ ax_n^2+2bx_n+c=2x_1\\ \end{matrix}
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UP 0 DOWN 0 1 6
6 Answers
21
131)x_{101} = 1^2-2^2+3^3-4^2+...+ 99^2-100^2+101^2\\ \\ x_{101}= (1+2)(1-2)+(3+4)(3-4)+...+(99+100)(99-100) +101^2
S = -(1+2+3+4+....+99+100) +101^2\\ \\ \ \ = -\frac{(100)(101)}{2} + 101^2 = 5151
341Q2: Consider f(x) = (2.6)^x-(2.4)^x -1
We can see that f(x)<-1 for x<0
When x≥0:
Consider
f'(x) = (2.6)^x \ln (2.6)-(2.4)^x \ln (2.4) > \ln (2.4) \left((2.6)^x-(2.4)^x \right)>0
Hence there is at most one root.
Since f(x) is continuous, x=2 is the only root
341Q3: Is quite familiar to me somehow:
The typical equation looks like ax_i^2 - 2bx_i +c = 2x_{i+1}
(Here we consider xn+1 = x1)
This can be written as ax_i^2 - 2(b-1)x_i +c = 2(x_{i+1}-x_i)
WLOG a>0. Then each term on LHS>0
Hence \sum_{i=1}^n ax_i^2 - 2(b-1)x_i +c >0
But \sum_{i=1}^n 2(x_{i+1}-x_i) = 0
Thus no real solutions are possible
