We have
$(n+1)^n - 1$
$= \sum_{k=0}^{n} {n \choose k} n^{n-k} -1 $
$= \sum_{k=0}^{n-2} {n \choose k} n^{n-k} + {n \choose n-1} n^{n-(n-1)} + {n \choose n} n^{n-n} -1$
$= \sum_{k=0}^{n-2} {n \choose k} n^{n-k} +n * n + 1 - 1$
$= \sum_{k=0}^{n-2} {n \choose k} n^{n-k} +n^2$
now each term in sum is having $n^2$ as a term and hence the expression is divisible by $n^2$
Because the number is not divsible by 2,3,5 for any n so let us check if is divisible by 7 for some n.
Now as 7 is co-prime to 10 so as per fermats little theorem
$10^6 \equiv 1 \pmod 7\cdots(1)$
By checking from1 to 6 we see that
$3^4 = 81 \equiv 4 \pmod 7$
or $10^4 = 81 \equiv 4 \pmod 7$ as $ 10\equiv 3 \pmod 7$
using (1) we get
$10^{6k+ 4} = \equiv 4 \pmod 7$
or $10^{6k+ 4} + 3 = \equiv 0 \pmod 7$
it is 27. because if we leave out 27 then none of the numbers above 27 is divisible by 27 so we shall not have a factor 27 of the LCM but for numbers 1 to 50 LCM shall have a factor 27.
27 is not by magic. it has to a prime number or a power of a prime. if it is composite other than power of a prime then is has got smaller factor and this is taken care of because smaller numbers are taken care of. so we should look for a number greater than 1/2 of the number because if a is taken care of the 2a is taken care of. in LCM. so we look through numbers 26,27 etc and find 27.
Let us consider 3 cases. Let the number be n
1) n is odd then 2 numbers are (2, n-1)
2) n is even and of the form 4n. Then the 2 numbers (2n-1, 2n+1)
3) n is even and of the form 4n +2. Then the 2 numbers (2n -1 , 2n + 3)
Out of 2n objects n objects can be chosen in
$\dfrac{(2n)!}{(n!)^2}$ ways
Now let us make 2n objects into 2 groups of n objects each.
For picking n objects from the set we need k objects from 1st set and n-k from 2nd set and n varies from 0 to n so number of ways
$\sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k}$
The 2 above are same as it shows the number of ways in 2 different ways
So
$\dfrac{(2n)!}{(n!)^2}= \sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k}$
Now as $\binom{n}{k} = \binom{n}{n-k}$ so we get the result
We have
$\sqrt{x^2+4x+13}+\sqrt{x^2-8x+41}$
= $\sqrt{(x+2)^2+9}+\sqrt{(x-4)^2 + 25}$
The 1st term that is $\sqrt{(x+2)^2+9}$ is distance from (x,0) to (-2,-3) and second term $\sqrt{(x-4)^2+25}$ is distance from (x,0) to (4,-5).
clearly sum of the distanace to (x,0) is lowest when (x,0), (-2,3) and (4,5) are in a straight line that is (x,0) is on the line from (-2,3) to (4,5) now as (x,0) lies in between the minimum is that distance from (-2.-3) to (4,5) or $\sqrt{(4+2)^2 + (5+3)^2} = \sqrt{100} = 10$
Because of symmetry if (x,y,z) is a solution then any permutation of (x,y,z) is also a solution
without loss of generality let us assume that $ x<=y<=z$
So $x+y+z <= 3z$
puttying in the given equation we get
$3z >= xyz$
or $3>=xy$
This gives the following set in (x,y)
(1,1) giving 2 + z = z which does not have a solution
(1,2) giving 3 + z = 2z or z = 3
(1,3) giving 4 + z = 3z giving z = 2 which is a contradiction as it should not be less than 3
so solution set (1,2,3) or a permutation of the same.
Let x = n + r where n is the integer part and r is the fractional part
we have
$\lfloor x \rfloor ^2 = 2(x+r) - 1$
so $n^2 = 2n -1$ when $ r < \frac{1}{2}$ or $n^2 = 2n$ and $ \frac{1}{2} \le r < 1$
$n^2 = 2n -1$ when $ r < \frac{1}{2}$
gives $n^2 - 2n + 1 = 0$ or $(n-1)^2 =0 $ or n= 1 giving $ 1 \le x < 1.5$
$n^2 = 2n$ and $\frac{1}{2} \le r < 1$
gives n = 0 or 2 giving $ .5 \le x < 1$ or giving $ 2.5 \le x < 3$
combining them we have $ .5 \le x < 1. 5 $ or $ 2.5 \le x < 3$
We are given
$\sum_{k=1}^{n} a_{2k-1}^2 = x\cdots(1)$
$\sum_{k=1}^{n} a_{2k}^2 = y\cdots(2)$
subtract (1) from (2) to get
$\sum_{k=1}^{n} (a_{2k}^2- a_{2k-1}^2) = y-x$
Or $\sum_{k=1}^{n} (a_{2k}- a_{2k-1})(a_{2k} + a_{2k-1}) = y-x$
But $(a_{2k}- a_{2k-1}= d$ common difference so we get
$\sum_{k=1}^{n} d(a_{2k} + a_{2k-1}) = y-x$
or $d \sum_{k=1}^{n} (a_{2k} + a_{2k-1}) = y-x$
Or $d \sum_{k=1}^{2n} (a_{k}) = y-x\cdots($
as $a_k = a_1 + (k-1) d$ for any k so we have
Now $a_k + a_{2n+1-k} = a_1 + (k-1)d + a_1 + (2n+1-k-1)d = 2a_1 + (2n-1) d = = a_1 + a_1 + (2n-1) d = a_1 + a_{2n}$
so $a_n + a_{n+1}d = a_1 + a_{2n} = z$
so $a_k + a_{2n+1-k} = z$
so
$d \sum_{k=1}^{2n} (a_{k}) $
$= d \sum_{k=1}^{n} (a_{k} + a_{2n+1-k})$
$= d \sum_{k=1}^{n} z$
= 2dnz
So $2dnz = y-x$
$P(x)=x^8-4x^7+7x^6+ax^5+bx^4+cx^3+dx^2+ex+f$ factorizes into eight linear factors $x-x_i$ with $x_i>0$ for $i=1,\,2,\,\cdots,\,8$.
Determine all possible values of $f$.
Solution
Using Vieta's formula we have
$\sum_{i=1}^8 x_i = 4\dots(1)$
$\sum_{i=1}^7 \sum_{j=i+1}^8 x_ix_j = 7\cdots(2)$
$\prod_{i=1}^8 x_i = f\cdots(3)$
We have
$\sum_{i=1}^8 x_i^2= (\sum_{i=1}^8 x_i)^2 - 2 \sum_{i=1}^7 \sum_{j=i+1}^8 x_ix_j $
$= 4^2 - 2 * 7 = 2$
or $\sum_{i=1}^8 x_i^2 = 2$
Subtracting (1) from above
$\sum_{i=1}^8 (x_i^2 - x_i) = -2$
adding $\frac{1}{4}$ to each term on LHS that is 2 and adding 2 on RHS we get
$\sum_{i=1}^8 (x_i^2 - x_i + \frac{1}{4}) = 0$
or $\sum_{i=1}^8 (x_i - \frac{1}{2})^2 = 0$
so $x_i = \frac{1}{2}$ for each i.
this satisfies the criteria that $x_i$ is positive
putting this in (3) we get $f = \frac{1}{\sqrt[8]2}$
We are given
$1= a+ b = a^{a+b} + b^{a+b}$
So $1- (a^ab^b + a^b b^a)$
$= a^{a+b} + b^{a+b} - (a^ab^b + a^b b^a)$
$= a^a(a^b-b^b) + b^a(b^b-a^b) = (a^a - b^a)(a^b - b^b)$
For a > b both the terms are non -ve so we have and if b > a then both terms are -ve and hence above is positive
$1- (a^ab^b + a^b b^a) >=0$ and hence the result
we are given $2^x = 3^y = 6^{-z}$
so $2 = 6^{\frac{-z}{x}}$
and $3 = 6^\frac{-z}{y}$
so $2 * 3 = 6^{(-\frac{z}{x} - \frac{z}{y})}$
or $6^1 = 6^{(-\frac{-z}{x} - \frac{z}{y})}$
or $1 = -\frac{z}{x} - \frac{z}{y}$
or $\frac{z}{x} +\frac{z}{y} + 1 = 0 $
or $\frac{1}{x} +\frac{1}{y} + \frac{1}{z}= 0$
Let the numbers be x and y
We have x+ y = 15
$(x+y)^3 = x^3 + y^3 + 3xy(x+y)$
or $15^3 = x^3 + y^3 + 3xy * 15$
or $x^3 + y^3 = 15^3 - 45xy$
this is mininum when xy is maximum
$x + y = 15$
we have $4xy = (x+y)^2 - (x-y)^2 = 15 - (-x-y)^2$
so xy is maximumum when x = y
or $x^3 + y^3$ is minumum when $x = y = 7.5$ and value is $2 * 7.5^3 = 843.75$
We have
ab = cd
or $\frac{a}{c}= \frac{d}{b} = \frac{m}{n}$ where m and n are in lowest terms or gcd(m,n) = 1
so an = cm and dn = bm
now $n^2(a^2+b^2+ c^2+d^2) $
$= (na)^2 + (nb)^2 + (nc)^2 + (nd)^2$
$= (cm)^2 + (bn)^2 +(nc)^2 + (nd)^2= (b^2+c^2)(m^2 + n^2)$
or $a^2+b^2+c^2+d^2 = \frac{b^2+c^2}{n^2} (m^2+n^2)$
as n is less than $b^2+c^2$ so it is product of 2 numbers > 1 so composite
We have p+q + r = 0
So $p^3 + q^3 + r^3 = 3pqr\cdots(1)$
And p+q = -r
or $(p+q)^2 = r^2\cdots(2)$
Similary $(q+r)^2 = p^2\cdots(3)$
$(r+p)^2 = q^2\cdots(4)$
Hence $\frac{(p+q)^2}{3pq}+\frac{(q+r)^2}{3qr} +\frac{(r+p)^2}{3rp}$
$= \frac{r^2}{3pq}+\frac{p^2}{3qr} +\frac{q^2}{3rp}$ (from (2), (3), (4)
$= \frac{r^3+q^3+ p^3}{3pqr} = \frac{3pqr}{3pqr} = 1$ (using (1))
We shall prove the same for 2 cases
1) n is even
2) n is odd
let us 1st prove for n even
case 1 :For n even say 2k $(>=2)$
$3^n + 1 = 3^{2k} + 1 = 9^k + 1 \equiv 2 \pmod 4$
so $3^n + 1$ is not divisible by 4 so cannot be divisible by $2^n$
case 2: for n odd say 2k + 1 $( >=3)$
$3^n + 1 = 3^{2k+ 1} + 3 = 9^k.3 + 1 \equiv 4 \pmod 8$
so so $3^n + 1$ is not divisible by 8 so cannot be divisible by $2^n$
We are given
$x+y+z = 3 \cdots(1)$
now $(x-1)^2\ge 0$
or $x^2-2x + 1 \ge 0$
or $x^2 \ge 2x -1\cdots(2)$
similarly $y^2 \ge 2y-1\cdots(3)$
and $z^2\ge 2z -1\cdots(4)$
addding (2), (3), (4) we get $x^2+y^2 + z^2\ge 2(x+y+z) - 3$
or $x^2+y^2 + z^2\ge 2* 3- 3$ (from (1)
or $x^2+y^2+z^2 \ge 3$
