@naga25french: In the stair case situation shalll the series start with first term as 1 2 3 and onwards?
@naga25french said:
Tweaked version of it :You have staircase with 10 steps . You can climbs either one or two steps at a time .. In how many ways you can reach the top of staircase ?
If we have 1 Stair => 1 way. [2nd Fibonacci Number F(2) ]
If 2 Stairs are there => 2 ways. [ F(3) ]
If 3 Stairs are there => 3 ways. => F(4)
So, For n Stairs, F(n+1) ways are there.
For 10 Stairs, F(11) = 89 ways. :roll:
@saurav5517 said:Now if the first toss is tails , second toss can be heads ..So tails can occur from 3rd toss and heads from the second toss..Shall the logic be the same??
yes .. be it head or tail , your series will 2,3
stair prob : Yes .. it will start from 1.2 .. so tenth number = 89
stair prob : Yes .. it will start from 1.2 .. so tenth number = 89
@naga25french said:yes .. be it head or tail , your series will 2,3
stair prob : Yes .. it will start from 1.2 .. so tenth number = 89
Got it buddy...:)
PS: I am not so great in algebra but your explanation was ultimate..Thanks bro...
If f(x) + f(x+1) = 2x^2 and f(33) = 99, find f(99).
@Estallar12 said:If f(x) + f(x+1) = 2x^2 and f(33) = 99, find f(99).
f(34)=2*33^2-99
f(35)=2*34^2-2*33^2+99=2(34+33) +99
..
..
f(99)=2(98^2-97^2+96^2-95^2 +.....+34^2-33^2)+99
f(99)=2(98+97+,,,,,,+33)+99
f(99)=2*33(131)+99
f(99)=33(262+3)
f(99)=265*33=2915*3=8745??
what are the options??
EDIT @Estallar12
@Estallar12 said:If f(x) + f(x+1) = 2x^2 and f(33) = 99, find f(99).
f(x+1)=2x^2-f(x)
f(34)=2*33^2-99
f(35)=2*34^2-2*33^2+99
f(36)=2*35^2-2*34^2+2*33^2-99
.
f(34)=2*33^2-99
f(35)=2*34^2-2*33^2+99
f(36)=2*35^2-2*34^2+2*33^2-99
.
.
f(99)=2*(98^2-97^2+96^2-95^2+......................+99
f(99)=2*(98+97+96+96+...............34+33)+99
f(99)=66*(2*33+65)+99
f(99)=2*(98+97+96+96+...............34+33)+99
f(99)=66*(2*33+65)+99
f(99)=8745
Q1
@vbhvgupta said:Q1
take the 7th term as "a" and the c.r= r
so the nos will be: (a/r^6), (a/r^5),(a/r^4)...a, ar, ar^2, ar^3...ar^6
their product: a^13= 2^13= 8192
Q2
the roots of x^3 - px^2+ qx - r=0 are consecutive integers . if q has the minimum possible value , which of the following is true
a) sum of roots is 0
b) produt of roots is -1
c) q= -1
d) exactly two of a, b, c.
how many non real roots does the eqn have if it has 1 negative and 2 positive roots
x^5-9x^4+13x^3+57x^2-86x-120=0
if f(x)= x^4 - 9x^3 + 26x^2 - 24x + 5 then one root of f(x) lies btwn
a)-1 &0
b)0&2
c)2&3
d)4&5
@raopradeep said:if f(x)= x^4 - 9x^3 + 26x^2 - 24x + 5 then one root of f(x) lies btwna)-1 &0b)0&2c)2&3d)4&5
d)4&5
f(4)=-5 and f(5)=35
Since there is a sign change between 4 and 5 so the graph would have crossed at least once between 4 & 5. So there is at least one root between 4 and 5.