Question

if 1+x^4+x^5=sigma(i=0 to 5)a sub i(1+x)^i, for all x in R,then asub2=?

Answer 1

how to go about this problem?

Answer 2

\[1+x^4+x^5\\ =a_0+a_1(1+x)+a_2(1+x)^2+a_3(1+x)^3+a_4(1+x)^4+a_5(1+x)^5\]
long way is to expand that ugly thing and combine like terms and then compare coefficients on both sides...
let me think about a short way...

Answer 3

ya i thought that would take a lot of time..and didn't proceed

Answer 4

Was there a time limit? Was this on a test?

Answer 5

yes

Answer 6

Here is one method. Let \(1 + x=t\).\[1 + (t-1)^4 + (t-1)^5 = \sum_{i=0}^5a_it^i\]Easier to expand the left side.

Answer 7

i think you can get the answer by differentiating 3 times
and then setting x to -1

Answer 8

This is high school stuff? o-o

Answer 9

what grade?

Answer 10

oh wow
yeah I got what parth would get

Answer 11

@ParthKohli even now its a bit too long...but i'll try this way

Answer 12

after all if both sides are the same function
then their derivatives should be equal
and then their double derivatives should be equal
and then their triple derivatives should be equal
so it makes sense

Answer 13

you will find parth thing easy to expand if you use pascal's triangle

Answer 14

i used binomial theorem

Answer 15

i m getting:\[1+t^5-4t^4+6t^3-4t^2+t\]

Answer 16

on the left side

Answer 17

that looks right

Answer 18

and just compare both sides to find the a_2 value

Answer 19

ooo, then we didn't even need to expand the entire thing.

Answer 20

what i did'nt get u?

Answer 21

there is actually a formula for finding binomial coefficients is what he is referring to

Answer 22

anyway it was a nice idea @ParthKohli

Answer 23

i got the answer:-4

Answer 24

my way for fun:
\[1+x^4+x^5 \\ = \\ a_0+a_1(1+x)+a_2(1+x)^2+a_3(1+x)^3+a_4(1+x)^4+a_5(1+x)^5 \\ \text{ differentiate both sides 3 \times } \\ \text{ should give } \\ 12x^2+20x^3=2a_2+6a_3(1+x)+12a_4(1+x)^2+20a_5(1+x)^3 \\ \text{ then replace } x=-1 \\ \\ 12(-1)^2+20(-1)=2a_2 \\ \text{ solve for } a_2 \]

Answer 25

nice :)

Answer 26

oh nice idea @freckles!

Answer 27

you nice job too
I didn't even think about doing the whole t=x+1 thing
and using binomial

Answer 28

thank u so much

Answer 29

i have 1 more q..

Answer 30

can any of u help?

Answer 31

what's the question

Answer 32

f(n)=\[[1/3 + 3n/100]n where [n] denotes the gratest integer less than or equal \to n.then \sum_{n=1}^{56} f(n) is ??\]

Answer 33

can u view the full q?

Answer 34

are you preparing for jee

Answer 35

\[f(n) = n \lfloor \frac{1}3 + \frac{3n}{100}\rfloor \]This is the function, right?

Answer 36

yes

Answer 37

can u read the full Q?

Answer 38

Break the bracket part of it (floor/greatest integer function) into pieces:

if \(5/3 > 3n/100\ge 2/3 \) then it is 1.
and so on.

Answer 39

ya i kind of did it like that but it took some time.. and i had to check like this for each term..

Answer 40

is it the proper way?

Answer 41

yes, this is the proper way. just find the intervals where it is zero, one, two and so on.

Answer 42

like that for all the 56 terms?

Answer 43

oh, you can't afford to plug in each and every number. just find the \(ranges\).

Answer 44

well..what i did was:
i found the ranges ..
like..till n=24 the value inside the braket is going to remain<1 right?

Answer 45

then till n=57 the value inside the bracket is gonna be <2..

Answer 46

then i did n(n+1)/2 till n=24 to sum them up

Answer 47

that is 300

Answer 48

then i added from 25,26,......,56 using sum of an A.P... i got 800

Answer 49

so on the whole 1100..but thats not the answer

Answer 50

where did i go wrong?

Answer 51

@freckles ? @ParthKohli ?

Answer 52

\[n \le \lfloor x \rfloor <n+1 \\ n \text{ is an integer } \\ \text{ you only need 3 intervals \to look at } \\ 0 \le \frac{100+9n}{300} <1 \text{ then } \lfloor \frac{100+9n}{300} \rfloor=0 \\ 1 \le \frac{100+9n}{300}<2 \text{ then } \lfloor \frac{100+9n}{300} \rfloor=1 \\ 2 \le \frac{100+9n}{300} <3 \text{ then } \lfloor \frac{100+9n}{300} \rfloor=2\]
solve each for n
I'm not sure have you done this?

Answer 53

oops i didn't finish my first line
\[n \le \lfloor x \rfloor \le n+1 \text{ then } \lfloor x \rfloor=n\]

Answer 54

i got just 2 intervals 0 to 1 and 1 to 2 considering n can take max 56 only...

Answer 55

example first inequality
\[0 \le \frac{100+9n}{300} <1 \\ 0 \le 100+9n<300 \\ -100 \le 9n <200 \\ \frac{-100}{9} \le n <\frac{200}{9} \\ \\ \text{ intersect with } 1 \le n \le 56 \\ \text{ we have } 1 \le n \le 22 \\ \text{ so } n \in [1,22] \text{ then } \lfloor \frac{100+9n}{300} \rfloor=0\]

Answer 56

and so on...

Answer 57

oh i stoped at 24...but i must stop at 22..thanks

Answer 58

right you only want to include integer values of n between -100/9 and 200/9
that intersect with integer values in [1,56]

Answer 59

but then you have two more intervals to consider
(I say two more as a hint but this should become clear after you do the solving of the inequalities above)

Answer 60

I'm going to go asleep now
it is 1:43 am for me

Answer 61

unless you have a question about this question

Answer 62

ok sorry to have disturbed u

Answer 63

thank u so much

Answer 64

no please don't say that
you didn't
I like math
it is just late and I'm old
old people get sleepy
and they have to work

Answer 65

i liked your questions
they were fun
(I don't say that about all the questions on here :p)

Answer 66

thank u

Answer 67

so are you good on this problem?

Answer 68

well....i found the intervals..by ur method...[1,22],[22,55] and 56

Answer 69

we can leave [1,22] coz inside the bracket we will have zero

Answer 70

so i added the A.P 23,24.....,55

Answer 71

[1,22]
[23,55]
{56}

Answer 72

just corrected that one 22 to 23

Answer 73

then atlast i added this to 2*56

Answer 74

is that correct?

Answer 75

you have 3 terms
the last term is 2*56

Answer 76

the sum of A.P=1343

Answer 77

I do not get that number

Answer 78

1343+(2*54) is this the answer?

Answer 79

I don't get that either

Answer 80

didn't you say [1,22] we have that floor part is 0
and [23,55] we have that floor part is 1
and then for {56} we have that floor part is 2
correct?

Answer 81

\[\sum_{i=1}^{22} i \cdot 0 + \sum_{i=23}^{55} i \cdot 1 +56 \cdot 2 \]

Answer 82

that first term is just 0

Answer 83

ya thats what i did

Answer 84

final answer:1455 (this is what i got)but answer:1399

Answer 85

1399 is correct

Answer 86

I think you aren't computing middle term right

Answer 87

i did compute..maybe i made a mistake in adding from 23 to 55 using ap..

Answer 88

\[\sum_{i=23}^{55} i \\ \sum_{i-22=1}^{i=55} i \\ \\ \text{ \let } u=i-22 \\ \\ i=55 \text{ then } u=55-22=33 \\ \text{ so we have } \\ \sum_{u=1}^{33} (u+22) =\frac{33(33+1)}{2}+22(33)\]

Answer 89

i got it!!i had made a silly mistake in calculation

Answer 90

Thanks a lot!!

Answer 91

np
you guys have fun:)