2017/011) If the sum of first n terms of an A.P. is $cn^2$ find the sum of squares of these n terms.

the sum of $1^{st}$ n terms is $cn^2$
so $k^{th}$ term is $ck^2-c(k-1)^2= c*(2k-1)$
so sum of n terms = $\sum_{k=1}^{n} (c^2(2k-1)^2) = c^2 \sum_{k=1}^{n} (4k^2-4k + 1) = c^2(4\frac{n(n+1)(2n+1)}{6}-4 \frac{n(n+1)}{2} +n)$
= $\frac{nc^2(4n^2-1)}{3}$

2017/010) Let $a,\,b,\,c,\,d$ be real with condition that $a+b\sqrt{2}+c\sqrt{3}+2d >= \sqrt{10(a^2+b^2+c^2+d^2)}$

show that $a^2+d^2=b^2+c^2$

Proof:

by Cauchy-Schawartz in equality we have
$(a^2+b^2+c^2+d^2)(1^2 + \sqrt2^2+ \sqrt3^2+2^2)>=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$
Or $10(a^2 + b^2 + c^2+ d^2)>=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$
Or $\sqrt{10(a^2 + b^2+ c^2+d^2)}>=(a+b\sqrt{2}+c\sqrt{3}+2d)$
from given condition and above we have
$\sqrt{10(a^2 + b^2+c^2+d^2)}=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$
when
$\frac{a}{1}= \frac{b}{\sqrt2}=\frac{c}{\sqrt3}= \frac{d}{2}= k (say)$
So $a = k, b^2= 2k^2,c^2=3k^2,d=2k$ and hence $a^2+d^2=b^2+c^2$

2017/009}Prove that $\sqrt{\frac{1\cdot 2}{3^2}}+\sqrt{\frac{2\cdot 3}{5^2}}+\sqrt{\frac{3\cdot 4}{7^2}}+\cdots+\sqrt{\frac{4032\cdot 4033}{8065^2}}\lt 2016$

we have $n^{th}$ term = $\frac{\sqrt{n\cdot (n+1)}}{2n+1}$
$= \frac{\sqrt{n^2+n}}{2n+1}$
$= \frac{\sqrt{n^2+n+\frac{1}{4}-\frac{1}{4}}}{2n+1}$
$=  \frac{\sqrt{(n+\frac{1}{2})^2-\frac{1}{4}}}{2n+1}$
$ < \frac{n+\frac{1}{2}}{2n+1}$
$ < \frac{1}{2}$
each term is $ < \frac{1}{2}$ and there are 4032 terms so sum is less than 2016

2017/008) Let P be a polynomial such that $P(x)=a_0+a_1x+?+a_nx^n$where $a_0,a_1,\cdots$? are non-negative integer. Given that P(1)=4 and P(5)=152 find P(6)

P(x) is a cubic polynomial as P(5) is 152 between $125(5^3)$ and $625(5^4)$
now coefficient of $x^3$ is 1
so $P(x) = x^3 + ax^2 + bx + c$
taking mod 5 we get c =2 so $P(X) = x^3 + ax^2+ bx + 2$ and so $a . 5^2 + 5b = 152-125-2 = 27$ giving a = 1, b= 2
sp $P(x) = x^3 +x ^2 + 2$ and it satisfies P(1) = 4
so P(6) = 254

2017/007) if $\alpha,\beta$ are the roots of $x^2-5x+1$ show that $\alpha^n+\beta^n$ is an integer and not divisible by 4 for any integer n

we have for n = 1 $\alpha+\beta = 5 $ which is integer not divsible by 4
and $\alpha\beta=1$
hence $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 5^2 - 2 = 23$ integer and not divisible by 4
hence it is true for n =1 and 2
let it be true for n = 1 to k for k >=2
now
$( \alpha^n+\beta^n)(\alpha+\beta) = \alpha^{n+1}+\beta^{n+1} + (\alpha^{n-1} + \beta^{n-1} \alpha\beta$
or $\alpha^{n+1}+\beta^{n+1} = 5 (\alpha^n+\beta^n) - (\alpha^{n-1} + \beta^{n-1}$
or $\alpha^{n+1}+\beta^{n+1} =  4(\alpha^n+\beta^n) +  (\alpha^n+\beta^n)- (\alpha^{n-1} + \beta^{n-1} \alpha\beta$
Now let n be smallest integer so that $\alpha^{n}+\beta^{n}$ is divisible by 4
so $\alpha^{n-1}+\beta^{n-1} - \alpha^{n-2}+\beta^{n-2}$ is divsible by 4
so $\alpha^{n-3}+\beta^{n-3} = alpha^{n-1}+\beta^{n-1} - \alpha^{n-2}+\beta^{n-2} - 4alpha^{n-2}+\beta^{n-2}$
is divisible by 4 which is contadiction.
hence proved

2017/006) if $x= a + b $ $y = aw+bw^2$ and $z= aw^2+ bw$ then show that $x^3+y^3 + z^3 = 3(a^3+b^3)$ and $x^2+y^2 + z^2 = 6ab$

We have $x+y+z = a(1+w+w^2) + b(1+w+w^2) = 0$
so x^3 + y^3+ z^3 = 3xyz = 3(a+b)(aw+bw^2)(aw^2+bw) = 3(a+b)(a^2 w^3 + abw^2 + abw^4 + b^2w^3)$
= 3(a+b)(a^2 + ab(w^2+ w^4) + b^2) = 3(a+b)(a^2 + ab(w^2+w) + b^2)$
$ = 3(a+b)(a^2 - ab + b^) = 3(a^3+b^3)$
Further
$x^2 = a^2 + b^2 + 2ab$
$y^2 = a^2w^2 + b^2w + 2ab$
$z^2 = a^2w + b^2w^2 + 2ab$
hence $z^2+y^2+z ^2 = a^2(1+w+w^2) + b^2(1+w+w^2) + 6ab = 6ab$

2017/005) Solve $|z+1| = z+ 2 + 2|$

because LHS is real so z + 2i is real and let z + 2i = x or z = x - 2i
so |x+1 - 2i |= x +2|
or $(x+1)^2 + 4 = (x+2)^2$
or $x^2 + 2x + 5 = x^2 + 4x + 4$ or $x = \frac{1}{2}$
$z = \frac{1}{2} - 2i$