We shall prove it by taking the GCD
$GCD((n+1)!+1, n!+1)$
$= GCD((n+1)!+1- (n!+1), n!+1)$ using GCD(a,b) = GCD(a-mb,b) for any integer m
$= GCD((n+1)!- n!, n!+1)$
$= GCD((n!(n+1-1), n!+1)$
$= GCD(n!.n, n!+1)$
$= GCD(n!, n!+1)$ we can devide 1st term by n and GCD shall not change and 2nd term is not divisible by n
$= GCD( n!, n!+1-n)$ using GCD(a,b) = GCD(a-mb,b) for any integer m
$=GCD(n!,1)= 1$
So these are relativvly primes
We have $a+b+c = 0$
Hence $ a+ b = -c $
Squaring both sides $a^2+2ab + b^2 = c^2$
Adding $a^2+b^2$ on both sides we get
$2(a^2+ab+b^2) = a^2 + b^2 + c^2$
Or $a^2+ab + b^2 = \frac{1}{2}(a^2+b^2+c^2)$
So $\frac{ab}{a^2+ab + b^2} = \frac{2ab}{a^2+b^2+c^2}\cdots(1)$
Similarly we have $\frac{bc}{b^2+bc + c^2} = \frac{2bc}{a^2+b^2+c^2}\cdots(2)$
And $\frac{ca}{c^2+ca + a^2} = \frac{2ca}{a^2+b^2+c^2}\cdots(3)$
Adding (1) (2) and (3) we get
$\frac{ab}{a^2+ab+b^2} + \frac{bc}{b^2 + bc+c^2} + \frac{ca}{c^2 + ca + b^2}= \frac{2ab+2bc+2ca}{a^2+b^2+c^2}\cdots(4)$
Now staring with $a+b+c=0$ squaring both sides we get
$a^2+b^2+c^2 + 2ab + 2bc+2ca= 0$
Or $a^2+b^2 + c^2 = - (2ab+2bc+2ca)$
Or $\frac{2ab+2bc+2ca}{a^2 +b^2+ c^2} = -1\cdots(5)$
Form (4) and (5) we get $\frac{ab}{a^2+ab+b^2} + \frac{bc}{b^2 + bc+c^2} + \frac{ca}{c^2 + ca + b^2}= - 1$
First we note that $gcd(x,x+1) = 1$
$gcd(x, x^2 + x + 1) = GCD(x, x(x+1)) = 1$
and $gcd(x+1, x^2 + x + 1) = GCD(x+1, x(x+1)) = 1$
So $x,x+1,x^2+x+1$ are pairwise co-prime and to prove that the given expression is divisible by $x(x+1)(x^2+x+1)$ we need to show that it is divisible by $x$, $x+1$ and $x^2+x+1$
Now Let $P(x) = (x+1)^7−x^7−1$
To show that it is dvisible by x we need to show P(0) = 0
$P(0) = 1^7 - 0^7 -1 = 0$ so it is divisibe by x
To show that it is dvisible by x+1 we need to show P(-1) = 0
$P(-1) = (-1+1)^7 - (-1)7 -1 = 0$ so it is divisibe by x+1
Now we need to show that it is divisible by $x^2+x+1$
That is if $x^2+x+1=0$ then $P(x) = 0$
Because
$x^2+x+1=0\cdots(1)$ we get
$x+1 = - x^2\cdots(2)$
And multiplying by $x-1$ we have $x^3-1 = 0$
Or $x^3 = 1\cdots(3)$
Now $P(x) = (x+1)^7−x^7−1$
$= (-x^2)^7 - x^7 -1$ using (1)
$=x^{14} - x^7 -1$
$= -(x^3)^4 * x^2 - (x^3)^2 * x - 1$ using (2)
$= -x^2 - x - 1$
$= - (x^2+x+1)=0$ using (1)
So P(x) =0 if $x^2+x+1=0$
Hence $x^2+x+1$ is a factor
As $x,x+1,x^2+x+1$ are factors and are pairwise co-prime so $(x+1)^7−x^7−1$ is divisible by $x(x+1)(x^2+x+1)$
We have $n^4 -1 = (n^2+1)(n^2–1) = (n^2–4 + 5)(n^2–1) = (n^2–4)(n^2–1) + 5(n^2–1)$
$= (n+2)(n-2)(n+1)(n-1) + 5(n^2–1)$
now the 2nd term is multiple of 5 and 1st term is product of 4 numbers which along with are 5 consecutive numbers. so one of them has to be divisible by 5. but as n is not divisible by 5 so one of the 4 other numbers is a multiple of 5 and hence the product. as given number is sum of 2 numbers each multiple of 5 and hence the given expression.
We have $\frac{n^7}{7} + \frac{n^{13}}{13} + \frac{71n}{91}$
$=\frac{n^7-n +n }{7} + \frac{n^{13}-n + n}{13} + \frac{71n}{91}$
$=\frac{n^7-n} {7} + \frac{n^{13}-n}{13} + \frac{n} {7}+ \frac{n}{13} + \frac{71n}{91}$
$=\frac{n(n^6-1)}{7} + \frac{n(n^{12}-n) }{13} + n$
If n is divisble by 7 1st term is integer and if n is not divisibl by 7 $n^6-1$ is divsible by 7 as per Fermats Little Theorem is 1st term is integer similarly the 2nd term and 3rd term is integer for any integer n and hence the given expression
Let us have a look at the $t^{th}$ term.
Let us look at difference for a couple of terms
$t_2 = t_1 = 5- 1 = 4$
$t_3-t_2 = 10 -5 = 5$
$t_4-t_3 = 16 -10 = 6$
From the above we see that $t_k = t_{k-1} + k + 2$
From this let us find $t_n$ in terms of n
We have $t_n = 1 + \sum_{k=2}^{n} (k + 2)$
$= 1 + \frac{n(n+1)}{2} - 1 + 2(n-1)$
$= \frac{n^2}{2} + \frac{5n }{2} - 2$
So sum $S_n = \sum_{k=1}^n (\frac{k^2}{2} + \frac{5k }{2} - 2)$
$= \sum_{k=1}^n (\frac{k^2}{2}) + \sum_{k=1}^n\frac{5k }{2} - \sum_{k=1}^n2$
$=\frac{1}{2}(\frac{n(n+1)(2n+1)}{6} + \frac{5}{2} \frac{n(n+1)}{2} - 2n $
$= \frac{2n^3 + 3n^2 +n }{12} + \frac{5n^2 +5n}{4} - 2n$
$= \frac{2n^3 + 3n^2 + n + 15n^2 + 15n -24n}{12}$
$= \frac{2n^3 + 18n^2 -8n}{12}$
$= \frac{n^3 + 9n^2 -4n}{6}$
We have
$\sqrt{(x^2 + 4x + 13)}$ Completing the square we get
$= \sqrt{x^2+ 4x+ 4 + 9} = \sqrt{(x+2)^2 + 3^2}$
The above in a plane is the distance from $(-2,-3)$ to $(x,0)$
$\sqrt{(x^2 - 8x + 41)}$ Completing the square we get
$= \sqrt{x^2 - 8x+ 16 + 25} = \sqrt{(x-4)^2 + 5^2}$
The above in a plane is the distance from $(4,-5)$ to $(x,0)$ also from (4,5) to $(x,0)$ and because of symmetry as the line has to be to $(x,0)$ we can choose $(4,-5)$
The sum of the 2 expression is the distance from $(-2,3)$ to $(x,0)$ and from $(x,0)$ to $(4,5)$ or $(4,-5)$ as the 2 distances are same. it is better to chose a point on the other side of x axis so (4,5).
The distance ins minimum if $(-2,3),(x,0)$ and $(4,5)$ are in same line and so minimum distance is distance from $(-2,3)$ to $(4,5)$ $= \sqrt{(4+2)^2 + (5+3)^2} = 10$
Because it is a degree 6 polynomial so there are six roots let them be $x_1,x_2,x_3,x_4,x_5,x_6$
Because coefficient of $x^5$ is zero so sum of roots is zero
So $\sum_{i=1}^{6}x_i = 0\cdots(1)$
Because coefficient of $x^45$ is zero so double sum of roots is zero
So $\sum_{i=1}^{6}\sum_{j=1. j\ne i}^{6}x_i x_j = 0\cdots(2)$
Hence from (1) and (2) we have $\sum_{i=1}^{6}x_i^2 = 0$
If roots are real all $x_i$ are zero which is impossible if a,b,c,d are non zero
Hence there is a contradiction and all roots cannot be real
An odd number is of the form 2n+ 1, so let 3 odd numbers be 2a+1, 2b+1,2c+ 1
Now $(2a+1)^2 = 4a^2 + 4a + 1 = 4(a^2+ a) + 1$
$(2b+1)^2 = 4b^2 + 4b + 1 = 4(b^2+ b) + 1$
$(2c+1)^2 = 4c^2 + 4c + 1 = 4(c^2+ c) + 1$
We observe that square of an odd is of the from 4n + 1
Now $(2a+1)^2 + (2b+1)^2 + (2c + 1)^4 = 4(a^2 + a + b^2 + b + c^2 + c) + 3$
Above is an odd number and of the form 4n + 3 so cannot be a perfect square
So the answer is No
$log(10+3\sqrt{10}) + log(10+\sqrt{90+\sqrt{90}})+ log(10- \sqrt{90+\sqrt{90}})$
Solution:
Add the 2nd and 3rd expression and knowing $log\, a + log\, b = log\, ab$
$log(10+\sqrt{90+\sqrt{90}})+ log(10- \sqrt{90+\sqrt{90}})$
$= log((10+\sqrt{90+\sqrt{90}})(10-\sqrt{90+\sqrt{90}}))$
$= log(10^2 - (\sqrt{90 +\sqrt{90}})^2$ using $(a+b)(a-b) = a^ 2- b^2$
$=log( 100 - (90 + \sqrt{90}))$
$= log(100 - 90 - \sqrt{90})$
$= log (10 - \sqrt{3^2 * 10})$ factoring 90 to product of a square and no square
$= log( 10 - 3 \sqrt{10})$
So given expression = $log(10+3\sqrt{10}) + log( 10 - 3 \sqrt{10})$
$= log ( (10+3\sqrt{10})(10- 3\sqrt{10}))$ knowing $log\, a + log\, b = log\, ab$
$=log (10^2 - (3\sqrt{10})$
$= log (100 - 9 * 10)$
$=log\, 10 = 1$
We shall prove the same by principle of mathematical induction.
Let $P(n) = 6^{n+2}+7^{2n+1}$
We shall show that base step and induction step are true
Base step
Here we show that P(1) is divisible by 43
$P(1) = 6^ 3 + 7^3 = 216 + 343 = 559 = 43 * 13$ is divisible by 43
So base step is true
Now for induction step
Induction step
Let it to true k that is P(k) is divisible of 43
We shall show that P(k+1) is divisible by 43
P(k) is divisible by 43 so there exists an integer b such that
$P(k) = 6^{k+2}+7^{2k+1}= 43b$
We have $P(k+1) = 6^{(k+1)+2}+7^{2(k+1)+1}$
$= 6(6^{k+1}) + 49 * 7^{2k+1} $
$= 6(6^{k+1}) + (6+43) * 7^{2k+1} $
$= 6(6^{k+1}) + 6 * 7^{2k+1} + 43 * 7^{2k+1}$
$= 6(6^{k+1} + 7^{2k+1}) + 43 * 7^{2k+1}$
$= 6 * 43 b + 43 * 7^{2k+1}$
$= 43( 6b + 7^{2k+1})$
This is multiple of 43
As we have proved both the base step and induction step hence this is true hence proved
