Indian Statistical Institute B.Math & B.Stat : Integration

Let $n$ be positive integers and $[.]$ be the floor function. Evaluate the integral \[ \int_{0}^{n^{\frac{1}{k}}} [ x^k + n] dx\] Since \( 0<1<2<3< \dots < n-1 < n \) we have \( 0^{\frac{1}{k}}<1^{\frac{1}{k}}<2^{\frac{1}{k}}<3^{\frac{1}{k}}< \dots < (n-1)^{\frac{1}{k}} < n^{\frac{1}{k}} \) $$$$ Therefore \[ \int_{0}^{n^{\frac{1}{k}}} [ x^k + n] dx = \sum_{i=0}^{n-1} \int_{i^{\frac{1}{k}}}^{(i+1){\frac{1}{k}}} [x^k + n] dx \] Now for a fixed $i$ in the range, consider the integral \[\int_{i^{\frac{1}{k}}}^{(i+1)^{\frac{1}{k}}} [x^k + n] dx \] \[ i^{\frac{1}{k}} < x < (i+1)^{\frac{1}{k}} \implies i < x^k < i+1 \implies x^k = i + \epsilon \quad where \quad 0 < \epsilon < 1\] \[ therefore \quad [x^k +n] = [i + \epsilon+n]= [i+n+ \epsilon] = i+n \] \[ therefore \quad \int_{i^{\frac{1}{k}}}^{(i+1)^{\frac{1}{k}}} [x^k + n] dx = \int_{i^{\frac{1}{k}}}^{(i+1)^{\frac{1}{k}}} (i+n) dx = (i+n)\bigg((i+1)^{\frac{1}{k}}-i^{\frac{1}{k}}\bigg) \] \[ Thus \quad \int_{0}^{n^{\frac{1}{k}}} [ x^k + n] dx = \sum_{i=0}^{n-1} \int_{i^{\frac{1}{k}}}^{(i+1){\frac{1}{k}}} [x^k + n] dx = \sum_{i=0}^{n-1} (i+n)\bigg((i+1)^{\frac{1}{k}}-i^{\frac{1}{k}}\bigg) \] \[ = (n+0)\bigg(1^{\frac{1}{k}}-0^{\frac{1}{k}}\bigg) + (n+1)\bigg(2^{\frac{1}{k}}-1^{\frac{1}{k}}\bigg)+ \dots +(n+n-2)\bigg((n-1)^{\frac{1}{k}}-(n-2)^{\frac{1}{k}}\bigg) + (n+n-1)\bigg(n^{\frac{1}{k}}-(n-1)^{\frac{1}{k}}\bigg) \] \( = n \bigg( 1^{\frac{1}{k}}-0^{\frac{1}{k}} +2^{\frac{1}{k}}-1^{\frac{1}{k}}+\dots+(n-1)^{\frac{1}{k}}-(n-2)^{\frac{1}{k}}+n^{\frac{1}{k}}-(n-1)^{\frac{1}{k}} \bigg) \) \(+ 1\bigg(2^{\frac{1}{k}}-1^{\frac{1}{k}}\bigg)+2\bigg(3^{\frac{1}{k}}-2^{\frac{1}{k}}\bigg)\dots+(n-2)\bigg((n-1)^{\frac{1}{k}}-(n-2)^{\frac{1}{k}}\bigg)+(n-1)\bigg(n^{\frac{1}{k}}-(n-1)^{\frac{1}{k}}\bigg) \) $$$$ \( = n \times n^{\frac{1}{k}}+2^{\frac{1}{k}}-1^{\frac{1}{k}}+2 \times 3^{\frac{1}{k}}-2 \times 2^{\frac{1}{k}}+ \dots +(n-2)(n-1)^{\frac{1}{k}}-(n-2)(n-2)^{\frac{1}{k}}+ (n-1)n^{\frac{1}{k}}-(n-1)(n-1)^{\frac{1}{k}} \) $$$$ \( =n^{\frac{1+k}{k}}-1^{\frac{1}{k}}-2^{\frac{1}{k}}-\dots-(n-1)^{\frac{1}{k}}+(n-1)n^{\frac{1}{k}}\) $$$$ \( =n^{\frac{1+k}{k}}-1^{\frac{1}{k}}-2^{\frac{1}{k}}-\dots-(n-1)^{\frac{1}{k}}-n^{\frac{1}{k}}+n \times n^{\frac{1}{k}}\) $$$$ \( =n^{\frac{1+k}{k}}-1^{\frac{1}{k}}-2^{\frac{1}{k}}-\dots-(n-1)^{\frac{1}{k}}-n^{\frac{1}{k}}+n^{\frac{1+k}{k}}\) $$$$ \( =2n^{\frac{1+k}{k}}-1^{\frac{1}{k}}-2^{\frac{1}{k}}-\dots-(n-1)^{\frac{1}{k}}-n^{\frac{1}{k}}\) $$$$ \( = 2n^{\frac{1+k}{k}} - \sum_{i=1}^{n} i^\frac{1}{k} \)

Indian Statistical Institute B.Math & B.Stat : Inequality

Prove that \[ \frac{1}{2}. \frac{3}{4}.\frac{5}{6}. \frac{7}{8} \dots \frac{2n-1}{2n} < \frac{1}{\sqrt{2n+1}} \] Let \(a,b > 0 \quad and \quad a < b \), we first show that \( \frac{a}{b} < \frac{a+1}{b+1}\). Now \( \frac{a+1}{b+1} - \frac{a}{b} = \frac{b-a}{b(b+1)} > 0 \quad since \quad b-a,a,b >0\). $$$$ Using the above result we see that \[\frac{1}{2} < \frac{2}{3} \] \[\frac{3}{4} < \frac{4}{5} \] \[\dots\] \[ \frac{2n-1}{2n} < \frac{2n}{2n+1} \] Let \( x_n = \frac{1}{2}. \frac{3}{4}.\frac{5}{6}. \frac{7}{8} \dots \frac{2n-1}{2n} \quad and \quad y_n = \frac{2}{3}. \frac{4}{5}.\frac{6}{7}. \frac{8}{9} \dots \frac{2n}{2n+1} \). Using the above result it is easy to see that \( x_n < y_n \). $$$$ Now \[ x_n^2 = x_n . x_n < x_n .y_n = \frac{1}{2}. \frac{3}{4}.\frac{5}{6}. \frac{7}{8} \dots \frac{2n-1}{2n} . \frac{2}{3}. \frac{4}{5}.\frac{6}{7}. \frac{8}{9} \dots \frac{2n}{2n+1} = \frac{1}{2n+1} \] \[ \implies x_n < \frac{1}{\sqrt{2n+1}} \quad i.e., \quad \frac{1}{2}. \frac{3}{4}.\frac{5}{6}. \frac{7}{8} \dots \frac{2n-1}{2n} < \frac{1}{\sqrt{2n+1}}\]

Indian Statistical Institute B.Math & B.Stat : Differentiation

Let \( f: \mathbb{R} \to \mathbb{R} \) be a function that is differentiable $n+1$ times for some positive integer $n$. The $ith$ derivative of $f$ is denoted by $f^{(i)}$. Suppose \[ f(1)=f(0)=f^{(1)}(0)=\dots=f^{(n)}(0)=0.\] Prove that $f^{(n+1)}=0$ for some $x \in (0,1)$. $$$$ Since $f$ is $n+1$ times differentiable(everywhere in $\mathbb{R}$).\( f,f^{(1)},\dots,f^{(n)} \) are all differentiable and continuous in \( (0,1) \quad and \quad [0,1] \) respectively.$\dots (A)$ $$$$ Since \(f(0)=f(1)=0 \) By Rolles' Theorem $\exists \quad c_1$ in $(0,1)$ suct that $f^{(1)}(c_1)=0$ $$$$ Now consider the interval $[0,c_1]$. Note that \( f^{(1)}(0)=f^{(1)}(c_1)=0 \) and $f^{(1)}$ is continuous and differentiable in $[0,c_1]$ and $(0,c_1)$ respectively. (Using $\dots (A)$). Thus by Rolle's Theorem $\exists \quad c_2$ in $(0,c_1)$ suct that $f^{(2)}(c_2)=0$ $$$$ Now consider the interval $[0,c_2]$. Note that \( f^{(2)}(0)=f^{(2)}(c_2)=0 \) and $f^{(2)}$ is continuous and differentiable in $[0,c_2]$ and $(0,c_2)$ respectively. (Using $\dots (A)$). Thus by Rolle's Theorem $\exists \quad c_3$ in $(0,c_2)$ suct that $f^{(3)}(c_3)=0$ $$$$ Continuing like this, we get a point $c_{n}$ such that $f^{(n)}(c_{n})=0$ where $c_{n} \in (0,c_{n-1})$. $$$$ Now consider the interval $[0,c_n]$. Note that \( f^{(n)}(0)=f^{(n)}(c_n)=0 \) and $f^{(n)}$ is continuous and differentiable in $[0,c_n]$ and $(0,c_n)$ respectively. (Using $\dots (A)$). Thus by Rolle's Theorem $\exists \quad c_{n+1}$ in $(0,c_n)$ suct that $f^{(n+1)}(c_{n+1})=0$. Since \( 0 < c_{n+1} < c_{n} < \dots < c_1 < 1 \) thus $c_{n+1} \in (0,1)$ $$$$

Indian Statistical Institute B.Math & B.Stat: Integration and Continuity

If \( c \int_{0}^{1} xf(2x) dx = \int_{0}^{2} tf(t) dt,\) where $f$ is a positive continuous functions, then find then value of $c$. $$$$ In the R.H.S put $2x=t$, this gives \( c \int_{0}^{1} xf(2x) dx = 4\int_{0}^{1} xf(2x) dx = (c-4) \int_{0}^{1} xf(2x) dx =0 \implies c =4\) $$$$ Since $f$ is given to be a positive continuous function, $xf(x)$ is continuous and $>0 \in (0,1)$. Therefore \( \int_{0}^{1} xf(2x) dx > 0 \). This follows from the fact that if a continuous functions is positive at a point in its domain, the $\exists$ a open neighborhood containing the point and contained in the domain throughout which the function is positive.

Indian Statistical Institute B.Math & B.Stat : Continuity and Bijections

$Problem-1.$ Let \(f: \mathbb{R} \to \mathbb{R} \) be given by \(f(x) = x(x-1)(x+1) \). Then show that $f$ is onto but not 1-1. $$$$ The injectivity part is very much obvious. See that \( f(0)=f(1)=f(-1) = 0 \). Now for the surjectivity, observe that $f$ is an odd degree polynomial. Hence it must be surjective! (Think if not, then $\exists$ a real number which has no pre-image (say $r$ ). Now consider the polynomial \( g(x) = f(x)-r \)!! what else g(x) is also an odd degree polynomial with no real roots!!! ) $$$$ $Problem-2.$ Another Problem with a kick of continuity. Let \(f: \mathbb{R} \to \mathbb{R} \) be given by \(f(x) = x^3-3x^2+6x-5 \). Then show that $f$ is both onto and 1-1. $$$$ Again note that $f$ is an odd degree polynomial so it is surjective. Since $f$ is a polynomial it is continuous on $\mathbb{R}$. Also \(f'(x)=3x^2-6x+6 = 3(x^2-2x+2) = 3\{(x-1)^2 + 1 \} > 0 \) \( \forall x \in \mathbb{R}\). This shows that $f$ is strictly increasing. Thus $f$ is continuous and strictly increasing hence must be 1-1. $$$$ $Problem-3.$ Let \(f: \mathbb{R} \to \mathbb{R} \) be given by \(f(x) = x^2 - \frac{x^2}{1+x^2} \). Then show that $f$ is neither onto nor 1-1. $$$$ Clearly $f$ is not injective since \(f(x)=f(-x) \). Also \( f(x) = \frac{x^4}{1+x^2} \) which shows that $f$ assume non-negative values. Thus the \(Range_f \subset \mathbb{R} = Co-domain_f\). Thus $f$ is neither surjective. $$$$ $Problem-4.$ Let \( \phi :[0,1] \to [0,1] \) be a continuous and 1-1 function. Let \( \phi(0) = 0, \phi(1) = 1, \phi \big(\frac{1}{2}\big) = c, \phi \big(\frac{1}{4}\big) = d.\) Then show that $ c > d $. $$$$ This follows form the property of continuous functions. On an interval I ( not necessarily compact ), if a function is continuous and 1-1 it is strictly increasing. To prove this you can use the Intermediate Value Theorem.

Problems : Limits

Problems for Indian Statistical Institute, Chennai Mathematical Institute, JEE Main and Advanced. $$$$ \[1.Evaluate: \lim_{x \to \infty} \frac{20+2\sqrt{x}+3\sqrt[3]{x}}{2+\sqrt{4x-3}+\sqrt[3]{8x-4}}\] \[2.Evaluate: \lim_{x \to \infty} \big( x \sqrt{x^2+a^2}-\sqrt{x^4+a^4}\big)\] \[3.Evaluate: \lim_{x \to \infty} x^3 \big\{ \sqrt{x^2+\sqrt{x^4+1}}-\sqrt{2}x \big \}\] \[4.Evaluate: \lim_{x \to \infty} \sqrt{\frac{x-\cos^2 x}{x+\sin x}}\] \[5.Evaluate: \lim_{x \to \infty} [2\log(3x)-\log(x^2+1) ]\] 6. Let \( R_n =2+\sqrt{2+\sqrt{2+\dots+\sqrt{2}}}\) (n square roots signs). Then evaluate \(\lim_{n \to \infty} R_n \) $$$$ 7. If \(a_n = \bigg( 1+\frac{1}{n^2}\bigg)\bigg( 1+\frac{2^2}{n^2}\bigg)^2 \bigg( 1+\frac{3^2}{n^2}\bigg)^3 \dots \bigg( 1+\frac{n^2}{n^2}\bigg)^n \), then evaluate \[ \lim_{n \to \infty} a_n^{-\frac{1}{n^2}} \] $$$$ \[8.Evaluate \lim_{x \to \infty} \sqrt{x^2+x}-\sqrt{x^2+1}\] \[9. \lim_{x \to \frac{\pi}{2}} (\sin x)^{\tan x}\] \[10. \lim_{x \to 0} \frac{\cos x -1}{\sin^2 x}\] For PDF click here

Indian Statistical Institute B.Math & B.Stat : Roots of Polynomial

Let $c$ be a fixed real number. Show that a root of the equation \[x(x+1)(x+2)\dots(x+2009)=c\] can have multiplicity at most $2.$ $$$$ Let \( f(x) = x(x+1)(x+2)\dots(x+2009)-c \) $$$$ First we compute the derivative of $f(x)$ and see that \( f'(x) = (x+1)(x+2)\dots(x+2009)+x(x+2)\dots(x+2009)+\dots+x(x+1)\dots(x+r-1)(x+r+1)\dots(x+2009)+\) \(\dots+x(x+1)(x+2)\dots(x+2008) \) where $r$ is a positive integer less than $2009$. $$$$ Now \(f'(-r) = (-r)(-r+1)\dots(-1)(1)\dots(-r+2009) =(-1)^r r! (2009-r)! > 0\) $if$ $r$ is even, else $<0$.where \( r \in \{0,1,2,\dots,2008\}\) $$$$ Thus we have the following inequalities, \( f'(0) > 0, f'(-1) < 0, f'(-2) > 0, \dots, f'(2008) > 0, f'(2009) < 0 \) $$$$ This shows that $f'(x)=0$ has one real root in each of the intervals \( (-1,0),(-2,-1),\dots,(-2009,-2008) \). Since $degree$ of $f'(x)$ is $2009$, all the roots of $f'(x)=0$ is real and simple. Thus a root of $f'(x)=0$ cannot be a root of the equation $f''(x)=0$. So a root of $f(x)=0$ can have $multiplicity$ at most $2$. $$$$

Indian Statistical Institute B.Math & B.Stat : Integration

Let $n$ be a positive integer Define \[f(x) = min \{|x-1|,|x-2|,\dots,|x-n|\} \] $$$$ Then evaluate \[ \int_{0}^{n+1} f(x) dx \] $$$$ When \(0 < x < 1+ \frac{1}{2} \), $|x-1|=min \{|x-1|,|x-2|,\dots,|x-n|\}$. Now $|x-1|= 1-x$ for \( 0 < x < 1\) and $|x-1|= x-1 $ for $1 < x < 1+ \frac{1}{2}$ $$$$ So, \(\int_{0}^{1+\frac{1}{2}} f(x) dx = \int_{0}^{1} (1-x) dx + \int_{1}^{1+ \frac{1}{2}} (x-1) dx = \frac{1}{2}+ \frac{1}{2} \times \frac{1}{4} \dots (A)\) $$$$ When \(n+ \frac{1}{2} < x < n+1 \), $|x-n|=min \{|x-1|,|x-2|,\dots,|x-n|\}$. Now $|x-n|= x-n$ for \( n+ \frac{1}{2} < x < n+1 \)$$$$ So, \(\int_{n+\frac{1}{2}}^{n+1} f(x) dx = \int_{n+\frac{1}{2}}^{n+1} (x-n) dx = \frac{1}{2}- \frac{1}{2} \times \frac{1}{4} \dots (B)\) $$$$ Consider the diagram given below where $1 < k \leq n$,. When \( x \in \big(k-\frac{1}{2},k+\frac{1}{2} \big) \), $|x-k|$ is minimum among $|x-i|$ where $i=1,2,3,\dots,k-1,k+1,\dots,n$ $$$$ $$$$ So, \( \int_{k-\frac{1}{2}}^{k+\frac{1}{2}} f(x) dx = \int_{k-\frac{1}{2}}^{k+\frac{1}{2}} |x-k| dx = \int_{k-\frac{1}{2}}^{k} (k-x) dx + \int_{k}^{k+\frac{1}{2}} (x-k) dx = \frac{1}{4}\) for $k=2,3,4\dots,n$.$$$$ Summing for $k=2,3,4\dots,n$ we get \[\int_{1+\frac{1}{2}}^{n+\frac{1}{2}} f(x) dx = \frac{(n-1)}{4} \dots (C)\] $$$$ Adding equations $A,C$ and $B$ we get \[ \int_{0}^{n+1} f(x) dx = \int_{0}^{1+\frac{1}{2}} f(x) dx+ \int_{1+\frac{1}{2}}^{n+\frac{1}{2}} f(x) dx+\int_{n+\frac{1}{2}}^{n+1} f(x) dx = \frac{1}{2}+ \frac{1}{2} \times \frac{1}{4}+ \frac{(n-1)}{4}+ \frac{1}{2}- \frac{1}{2} \times \frac{1}{4} \] $$$$ \[= \frac{(n-1)}{4}+1 = \frac{n+3}{4} \]

Indian Statistical Institute B.Math & B.Stat : Square of a real is non-negative

Show that the following system of inequalities has exactly one solution $a-b^2 \geq \frac{1}{4},$ $b-c^2 \geq \frac{1}{4},$ $c-d^2 \geq \frac{1}{4}$ and $d-a^2 \geq \frac{1}{4}.$ $$$$ Adding up all the inequalities we get \( a-b^2 + b-c^2 + c-d^2 + d-a^2 \geq \frac{1}{4} + \frac{1}{4} + \frac{1}{4} +\frac{1}{4} \) $$$$ \( \implies a-a^2 - \frac{1}{4} + b-b^2- \frac{1}{4} + c-c^2 - \frac{1}{4} + d-d^2 - \frac{1}{4} \geq 0 \) $$$$ \( \implies -\big(a- \frac{1}{2} \big)^2 -\big(b- \frac{1}{2} \big)^2 - \big(c- \frac{1}{2} \big)^2 - \big(d- \frac{1}{2} \big)^2 \geq 0 \) $$$$ which is possible only when R.H.S is zero i.e., \( a=b=c=d= \frac{1}{2} \), since the R.H.S is always non-positive.

Indian Statistical Institute B.Math & B.Stat : Limits at Infinity

Let \( a_1 > a_2 > \dots > a_r \) be positive real numbers. Compute \(\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}}\). $$$$ Since \( a_1 > a_2 > \dots > a_r \) and each of them is positive we have \(a_1^n>a_2^n>\dots>a_r^n \) $$$$ \( \implies a_1^n+a_2^n+\dots+a_r^n < a_1^n+a_1^n+\dots+a_1^n = ra_1^n \) $$$$ Letting \( n \to \infty \) we have, \(\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} < \lim_{n \to \infty}(ra_1^n)^{\frac{1}{n}} \) \( = a_1\lim_{n \to \infty}r^{\frac{1}{n}}= a_1 \) Note $r>0$ $$$$ Now, \( \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} = \bigg( a_1^n \big(1+\frac{a_2^n}{a_1^n}+\dots+\frac{a_r}{a_1}\big) \bigg)^{\frac{1}{n}} = a_1\bigg( 1+\frac{a_2^n}{a_1^n}+\dots+\frac{a_r}{a_1}\big) \bigg)^{\frac{1}{n}} > a_1 \) Since \(\bigg( 1+\frac{a_2^n}{a_1^n}+\dots+\frac{a_r}{a_1} \bigg)^{\frac{1}{n}} > 1\) $$$$ Letting \( n \to \infty \) we have, \(\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} > \lim_{n \to \infty}a_1 =a_1 \) $$$$ Thus by $Sandwhich-theorem$ \(\lim_{n \to \infty} \big( a_1^n+a_2^n+\dots+a_r^n \big)^{\frac{1}{n}} = a_1\)

Inequality

$Problem$ #3 $$ $$ Find all real numbers $x$ for which \(\sqrt{3-x}-\sqrt{x+1} > \frac{1}{2}\) $$ $$ Let $f(x)$ \(=\sqrt{3-x}-\sqrt{x+1}\). First note that $f(x)$ is defined for \( -1 \leq x \leq 3 \) $$ $$ \( f'(x) = \frac{-1}{2\sqrt{3-x}} - \frac{1}{2\sqrt{x+1}} = - \big(\frac{1}{2\sqrt{3-x}} + \frac{1}{2\sqrt{x+1}}\big) < 0 \Rightarrow f(x)\) is strictly decreasing $$ $$ Now \( f(-1) = 2 > \frac{1}{2}\) and \( f(3) = -2 < \frac{1}{2}\) Since $f(x)$ is continuous, $\exists$ at least one x $\in$ ${(-1,3)}$ suct that $f(x) = \frac{1}{2}$ $$ $$ \( f(x) = \frac{1}{2} \Rightarrow \sqrt{3-x}-\sqrt{x+1} = \frac{1}{2} \Rightarrow 64x^2-128x+33 = 0 \Rightarrow x = 1 \pm \frac{\sqrt{31}}{8} \)$$ $$ but \(x = 1 + \frac{\sqrt{31}}{8}\) does not satisfy \(\sqrt{3-x}-\sqrt{x+1} = \frac{1}{2}\) Check yourself! So the only solution is \(x = 1 - \frac{\sqrt{31}}{8}\) $$ $$ Since $f(x)$ is strictly decreasing, the given inequality is true for \( x \in {[-1,1 - \frac{\sqrt{31}}{8}\big)}\) $$ $$

Matrices & Determinants

$Problem$ #1 $$ $$ If $A$ and $B$ are different matrices satisfying \( A^3 = B^3 \) and \(A^2B = B^2A\), find \(det(A^2+B^2)\) $$ $$ Since $A$ and $B$ are different matrices \( A-B \neq O \), Now \((A^2+B^2)(A-B) = A^3-A^2B+B^2A-B^3\) $$ $$ =$O$ since \(A^3 = B^3\) and \(A^2B = B^2A\) $$ $$ This shows that \((A^2+B^2)\) has a zero divisor, so it is not invertible hence \(det(A^2+B^2) = 0\)

Indian Statistical Institute B.Math & B.Stat : Number Theory

Let $k$ be any odd integer greater that 1. Then show that \( 1^k+2^k+3^k+\dots \dots+2006^k \) is divisible by 2013021. $$$$ We will prove the general case. Let \(S= 1^k+2^k+3^k+\dots \dots+n^k\) where $n \geq 2$ and \(n\in \mathbb{N}\). $$$$ \(2S = 1^k+2^k+3^k+\dots \dots+n^k + 1^k+2^k+3^k+\dots \dots+n^k = (1^k+n^k)+(2^k+(n-1)^k)+\dots \dots+(n^k+1^k)\) $$$$ Using the result, $n^k+m^k$ is always divisible by $n+m$ if $k$ is odd we see that $2S$ is divisible by $(n+1)$ $$$$ Again \(2S = 1^k+2^k+3^k+\dots \dots+n^k + 1^k+2^k+3^k+\dots \dots+n^k\) $$$$ \( = (1^k+(n-1)^k)+(2^k+(n-2)^k)+\dots \dots+((n-1)^k+1^k)+2n^k\) $$$$ Using the same result and noting that $2n^k$ is divisible by $n$ we see that $2S$ is divisible by $n$. Now both $n$ an $n+1$ divides $2S$ and $g.c.d(n.n+1)=1$ we see that $n(n+1)$ divides $2S$ this implies $\frac{n(n+1)}{2}$ divides $S$. ( $n(n+1)$ is always even) $$$$ In the given problem $n=2006$, thus \( 1^k+2^k+3^k+\dots \dots+2006^k \) is divisible by $\frac{2006(2006+1)}{2}=2013021$