using property,find the sum of the following:- 1,3,5,7,9,11,13,15,17,19,21.
Answers
Answer:
Sum of first n odd numbers=n²
here, n=11
therefore, answer= 11²=121
Answer:
121
Step-by-step explanation:
Then S = 1+3+5+7+9+11+13+15+17+19+21 ………………………………………(1)
Proof 1:
As the terms of series (1) are all odd numbers and there are 11 terms in the series, we can write
S = ∑(2n-1) , n=1,2,3,……..11
Or, S = ∑2n - ∑1 = 2∑n - (1+1+1+……n terms)
Now ∑n is nothing but the sum of the first natural numbers =n(n+1)/2 and summation of n numbers of 1 is n.
∴ S = 2n(n+1)/2 - n = n(n+1) - n = n² + n - n = n²
In the present problem, n = 11. Substituting for n
S = 11² = 121 (Proved)
Proof 2:
The terms 1,3,5, ………,21 of series (1) form an A.P. in which the first term (a) = 1, the common difference (d) = 2 and the number of terms (n) = 11,
∴ S = n/2 {2a + (n-1)d} = 11/2 {2x1 + (11–1)2} = 11/2 (2 + 10x2)
= (11 x 22)/2 = 11x11 = 11² = 121 (Proved)
Thus the sum of any number of consecutive odd numbers beginning with unity is square of the number of odd numbers.
Proof 3:
Writing the series (1) in reverse order,
S = 21+19+17+15+13+………………+1 ……………………………………..(2)
(1) + (2) →
2S = 22+22+22+………11 terms
∴ 2S = 22 x 11 = 2 x 11 x 11 Dividing both sides by 2,
S = 11 x 11 = 121 (Proved)
Proof 4:
Since the series is the sum of the first 11 positive odd numbers starting with unity or 1, we can directly apply the formula for the sum which is
S = n² = 11² = 121 (Proved)
Proof 5:
We can directly add the numbers in the series (1) manually or using a calculator.
This is the brute force method.
S = 1+3+5+7+9+11+13+15+17+19+21 = 4+5+7+9+11+13+15+17+19+21 = 9+7+9+11+13+15+17+19+21 = 16+9+11+13+15+17+19+21 = 25+11+13+15+17+19+21 = 36+13+15+17+19+21 = 49+15+17+19+21 = 64+17+19+21 = 81+19+21 = 100+21
= 121 (Proved)
the answer is always a square number.
More detail is provided:
find the largest number in the partial sum of odd consecutive integers starting at 1.
Add one to it and divide by two. Now you know how many terms there are.
square this number which is equal to the sum.
using the above example:
21 + 1 = 22
22 / 2 = 11
take the square of it: 11^2=121
It's 121. And it's pretty simple!
Let me explain with the mathematical concept of Arithmetic Progression.
If Sn is sum of n terms of an A.P. whose first term is “a” and last term is “l”,
then Sn = (n/2)(a + l)
Sum = (11/2)(1 + 21) = (11/2)(22) = 11 * 11 = 121
The sum of the first n numbers is 1+2+3+4+⋯=n(n+1)2 .
Double this and we get the sum of the first n even numbers: 2+4+6+8+⋯=n(n+1) .
The sum of the first 22 integers, by the initial formula, is 22⋅232=253 .
If we subtract off the first 11 even integers, we get the desired formula. We have 253 –11⋅12=253–132=121 .
This is mental arithmetic that you should be able to do in your head. If not, then use a calculator, entering each number in turn followed by the + plus except after the last number when you should enter the = sign. Just in case you can’t do either, the answer is 121.
The sum of first N odd numbers is N^2. Check the proof at The sum of the first n odd natural numbers.
1+3+5+7+9+11+13+15+17+19+21 = 11^2 = 121 as we are looking for sum of first 11 odd numbers.
121
1+3+5+7+9+11+13+15+17+19+21
Solution:
Here,
We see that this is a series of odd number & the total number of this series n=11
We know that the sum of the first n odd numbers is n^2.
So we get, 11^2=121
Answer is 121
When I add all the numbers up. I get 121. 21+19 is 40, plus 17 is 57, plus 15 is 72, plus 13 is 85, plus 11 is 96, plus 9 is 105, plus 5 is 110, plus 7+3+1 is a total of 11, so 110+11 is 121.
1 + 3 + 5 + 7 + 9 + (10 + 1) + (10 + 3) + (10 + 5) + (10 + 7) + (10 + 9) + 21
= (1 + 3 + 5 + 7 + 9) + (1 + 3 + 5 + 7 + 9) + 10 + 10 + 10 + 10 + 10+ 21
= (1 + 9) + (3 + 7) + (1 + 9) + (3 + 7) + (5 + 5) + 10 + 10 + 10 + 10 + 10 + 21
= 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 21
= 100 + 21
= 121