Q. 62 - Calculus

Question

Prove part (b) of Theorem 4.4 in the case when n is odd: If n is a positive odd integer, then $\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$(Hint: Use a method similar to the one for the previous exercise, but take note of what happens with the extra middle term of the sum.)

Step-by-Step Solution

Answer

 In the expanded form of sigma notation, the pair of first and last term is $n+1$and similarly, the sum of second and last-second terms will be also $n+1.$there are $\frac{n}{2}$pairs whose sum is $n+1$so the sum of sigma notation is equal to the sum of all those pairs$\frac{n(n+1)}{2}$.

 So $\sum _{k=1}^{n}k=\frac{n(n+1)}{2}.$

Step 1. Given information.

 The given formula is the following.

$\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$

Step 2. Proof.

Write the expanded form of sigma notation.

$\sum _{k=1}^{n}k=1+2+3+\cdots (n-1)+n$

The sum of the first and the last term is $1+n.$

the sum of the second and second last terms will be $2+(n-1)=n+1$

The number of pairs whose sum is $n+1$will be $\frac{n}{2}.$

So the sum of all pairs will be $\frac{n}{2}(n+1)=\frac{n(n+1)}{2}$

the sum of all pairs is equal to sigma notation.

so $\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$