Question 1

# Find the sum of odd integers from 1 to 2001.

The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

This sequence forms an A.P.

Here, first term, a = 1

Common difference, d = 2

Here,

\begin{align} a + (n - 1)d = 2001 \end{align}

\begin{align} => 1 + (n - 1)(2) = 2001 \end{align}

\begin{align} => 2n -2 = 2000 \end{align}

\begin{align} => n = 1001 \end{align}

\begin{align} S_n = \frac {n}{2}\left[2a + (n -1)d\right]\end{align}

\begin{align} \therefore S_n = \frac {1001}{2}\left[2 × 1 + (1001 -1)×2\right]\end{align}

\begin{align} = \frac {1001}{2}\left[2 + 1000×2\right]\end{align}

\begin{align} = \frac {1001}{2} × 2002\end{align}

\begin{align} =1001 × 1001 \end{align}

\begin{align} = 1002001 \end{align}

Thus, the sum of odd numbers from 1 to 2001 is 1002001.