Question

Find without adding: 1+3+5+7+9+11+13+15+17+19+21

Answer 1

Hint: Observe that the given series is an arithmetic series. Find the first term and the common difference and the number of the terms of the series. Use the fact that the sum of first n terms of an A.P is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Hence find the sum of the given series.

Complete step-by-step answer:
In the above series we observe that 3-1 = 5- 3 = 7-5 = …
Hence the given series forms an arithmetic series with the first term as 1 and common difference as 3 -1 =2
Also the number of terms of the series = 11.
We know that the sum of the first n terms of an AP $={{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Here a = 1, d = 2 and n = 11.
Hence we have ${{S}_{11}}=\dfrac{11}{2}\left[ 2\times 1+\left( 11-1 \right)2 \right]$
Taking 2 common from the terms inside the square braces, we get
${{S}_{11}}=\dfrac{11}{2}\times 2\left[ 1+10 \right]=11\times 11=121$
Hence the given series adds up to 121.

Note: Alternative solution:
Let S = 1+3+5+7+9+11+13+15+17+19+21
Writing S in reverse order, we get
S = 21+19+17+15+13+11+9+7+5+3+1
Adding, terms at respective positions with each other we get
2S = (21+1)+(19+3)+(17+5)+(15+7)+(13+9)+(11+11)+(9+13)+(7+15)+(5+17)+(3+19)+(1+21)
i.e. 2S = 22+22+22+22+22+22+22+22+22+22+22
Hence we have $2S=22\times 11=242$
Dividing by 2 on both sides, we get
$S=\dfrac{242}{2}=121$, which is the same as obtained above.