2016/006) show that 13 is the largest number that divides 2 numbers of the form $(n^2+3),(n+1)^2+3$

 let us find the GCD

$GCD(n^2+3,(n+1)^3+3) = GCD(n^2+3, n^2 + 2n + 4)$
$=GCD(n^2+3, n ^2+2n+4-(n^2+3))= GCD(n^2+3,2n+1)$
$=GCD(2*(n^2+3), 2n+ 1)$ we can multiply 1st one by 2 as 2nd one is odd
$=GCD(2n^2+6,2n++1)$
$=GCD(2n^2+6-n(2n+1),2n+1)$
$=GCD(6-n, 2n+ 1)$
$= GCD(12-2n,2n+1)$ we can multiply 1st one by 2 as 2nd one is odd
$= GCD(13,2n+1)$
So 13 is the largest number that divides both and when 2n+1 = 13 mod 13 or n= 6 mod 13

2016/005) Find sum $\sin^3x + \sin^3 (x+\frac{2\pi}{3}) + \sin^3 (x+\frac{4\pi}{3})$

$\sin^3x + \sin^3 (x+\frac{2\pi}{3}) + \sin^3 (x+\frac{4\pi}{3})$
$=\sin^3x + \sin^3 (x-\frac{\pi}{3}) + \sin^3 (x+\frac{\pi}{3})$
$= \frac{1}{4}(3\sin x - \sin 3x) + \frac{1}{4} (3\sin(x - \frac{\pi}{3}) - \sin(3x - \pi)) +   \frac{1}{4}(\sin(x + \frac{\pi}{3}) - \sin(3x + \pi))$
$= \frac{1}{4}(3\sin x - \sin 3x + 3\sin(x - \frac{\pi}{3} ) + \sin3x - 3\sin(x + \frac{\pi}{3}) + \sin 3x)$
$= (3/4) [\sin x - \sin 3x + \sin(x - \frac{\pi}{3}) + \sin(x + \frac{\pi}{3}]$
$=\frac{3}{4}(\sin x - \sin 3x - 2 \sin x \sin\frac{\pi}{6})$
$= \frac{3}{4}(\sin x - \sin3x - \sin x)$
$= - \frac{3}{4}\sin 3x$
The following has been used in proving the above identity

$1) \sin^3 x = \frac{3\sin x - \sin3x}{4}$
$2) \sin A + \sin B = 2\sin \frac{A+B}{2} \sin \frac{A-B}{2}$

2016/004) $(1+ni)^2$ is purely imaginary find n.

We have $(1+ni)^2 = 1 +2ni- n^2$
for it to be imaginary we need to have real part zero or $1-n^2 = 0$ or $n = \pm 1$

2016/003) In an AP sum of p terms is same as sum of q terms. Show that $(p+q)^{th}$ term is zero

let $1^{st}$ term be $a$ and difference $d$
let $k^{th}$ term be $t_k$ and sum upto k term be $S_k$
$t_{p}= a + (p - 1) d$
So $S_p= \frac{a + (p-1) d}{2} * p$
sum of q terms $S_q= \frac{a + (q- 1)d}{2}* q = \frac{a + (p-1) d}{2}* p$ (as both are same)
So $(a-d) q + dq ^2 = (a-d) p + dp^2$
or $(a-d)(q-p) = d(p^2 - q^2)$
so $(a-d) = -d (p+ q)$
or $a + (p+q-1) d = 0$ and this is the $(p+q)^{th}$ term and it is zero

2016/002) Let n be any positive integer $>1$ then show that $n^4+4^n$ is a composite number

there are 2 cases
1) n is even.
$n^4+4^n > 2$ and even so composite
2) n is odd
$n^4 + 4^n= n^4 + 2n^2*2^n + 4^n - 2n^2*2^n$
$=(n^2 + 2^n)^2 - 2n^2*2^n$
$=(n^2 + 2^n)^2 - n^2*2^{n+1}$
$= (n^2 + 2^n - n * 2^{\frac{n+1}{2}})(n^2 + 2^n + n * 2^{\frac{n+1}{2}})$
Therefore, if n is odd it has 2 factors so composite

So it is composite for any n

2016/001) How to find the nth term of a sequence whose consecutive terms difference' is in A.P?

for example $1,3,6,10,15\cdots$

Because difference is AP let us find the differences  $2,3,4,5$
$t_{n+1} - t_{n} = n$
so $t_n$ should be order 2 polynomial
say $t_n = an^2 + bn + c$
hence $t_{n+1}-t_n=a((n+1)^2-n^2) + b=2an +a +b = n$
so $2a = 1$ and $a+b = 1 => a = b = \frac{1}{2}$
putting $n = 1$ we get $c = 0$
so $t_n = \frac{1}{2}(n^2 + n) = \frac{1}{2}n(n+1)$