Express the following as a sum of consecutive odd numbers starting with 1.
1) ${{7}^{2}}$
2) ${{9}^{2}}$
3) ${{5}^{2}}$
4) ${{11}^{2}}$
Answer
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Hint:
Here we have to represent those numbers in terms of the sum of consecutive odd numbers. This will be an arithmetic series since the difference of any two consecutive terms is a constant. So we will use the formula of sum of arithmetic series here. From this formula we will get the number of terms of consecutive odd numbers and then we will express the given number in terms of the sum of consecutive odd numbers.
Complete step by step solution:
1) We know that the value of \[{7^2}\] is equal to 49.
We have to express 49 in terms of the sum of consecutive odd numbers starting with 1.
Here, sum of consecutive odd numbers starting with 1 will form an arithmetic progression i.e.
\[1 + 3 + 5 + 7 + ......n\] is an arithmetic progression as the difference between any two consecutive terms is a constant which is \[2\] here and the first term is \[1\] and let \[n\] be the number of terms.
As we know
Our series is:- \[1 + 3 + 5 + 7 + ......n\]
\[\begin{array}{l}a = 1\\d = 2\end{array}\]
Now putting values of \[a\] and\[d\] in the formula, we get;
\[{S_n} = \dfrac{n}{2}\left[ {2 \times 1 + \left( {n - 1} \right) \times 2} \right]\]
Simplifying it further, we get;
\[\begin{array}{l}{S_n} = \dfrac{n}{2}\left[ {2 + 2n - 2} \right]\\{S_n} = \dfrac{n}{2}\left[ {2n} \right]\\{S_n} = {n^2}\end{array}\]
As we know from the question that the value of sum of all terms is \[49\]
Now equating it with the obtained sum which is \[{n^2}\] .
\[49 = {n^2}\]
Taking square roots on both sides;
\[\begin{array}{l}\sqrt {49} = \sqrt {{n^2}} \\7 = n\end{array}\]
$\therefore n=7$
So the number of odd numbers is 7.
Now the expression of \[{7^2}\] in terms of the sum of consecutive odd numbers can be obtained as we have got the number of odd numbers needed to get the sum as 49.
So the required expression is
\[{7^2} = 1 + 3 + 5 + 7 + 9 + 11 + 13\]
2) \[{9^2}\]
Let’s first calculate the value of \[{9^2}\]which is equal to 81 .
We have to express \[81\] in terms of the sum of consecutive odd numbers starting with 1.
Here, sum of consecutive odd numbers starting with \[1\] will form an arithmetic progression i.e.
\[1 + 3 + 5 + 7 + ......n\] is an arithmetic progression as the difference between any two consecutive terms is a constant which is \[2\] here and the first term is \[1\] and let \[n\] be a number of terms.
Our series is:-\[1 + 3 + 5 + 7 + ......n\]
\[\begin{array}{l}a = 1\\d = 2\end{array}\]
Now putting values of \[a\] and\[d\] in the formula \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\], we get;
\[{S_n} = \dfrac{n}{2}\left[ {2 \times 1 + \left( {n - 1} \right) \times 2} \right]\]
Simplifying it further, we get;
\[\begin{array}{l}{S_n} = \dfrac{n}{2}\left[ {2 + 2n - 2} \right]\\{S_n} = \dfrac{n}{2}\left[ {2n} \right]\\{S_n} = {n^2}\end{array}\]
As we know from the question that the value of sum of all terms is \[81\]
Now equating it with the obtained sum which is \[{n^2}\] .
\[81 = {n^2}\]
Taking square roots on both sides;
\[\begin{array}{l}\sqrt {81} = \sqrt {{n^2}} \\9 = n\end{array}\]
$\therefore n=9$
So the number of odd numbers is \[9\] .
Now expression of \[{9^2}\]in terms of sum of consecutive odd numbers can be obtained as we have got the number of odd numbers needed to get the sum as \[81\]
So the required expression is
\[{9^2} = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17\]
3) \[{5^2}\]
Let’s first calculate the value of \[{5^2}\]which is equal to \[25\] .
We have to express \[25\]in terms of the sum of consecutive odd numbers starting with \[1\].
Here, sum of consecutive odd numbers starting with \[1\]will form an arithmetic progression i.e.
\[1 + 3 + 5 + 7 + ......n\] is an arithmetic progression as the difference between any two consecutive terms is a constant which is \[2\] here and the first term is \[1\] and let \[n\] be the number of terms.
Our series is:-\[1 + 3 + 5 + 7 + ......n\]
\[\begin{array}{l}a = 1\\d = 2\end{array}\]
Now putting values of \[a\] and\[d\] in the formula \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\], we get;
\[{S_n} = \dfrac{n}{2}\left[ {2 \times 1 + \left( {n - 1} \right) \times 2} \right]\]
Simplifying it further, we get;
\[\begin{array}{l}{S_n} = \dfrac{n}{2}\left[ {2 + 2n - 2} \right]\\{S_n} = \dfrac{n}{2}\left[ {2n} \right]\\{S_n} = {n^2}\end{array}\]
As we know from the question that the value of the sum of all terms is 25.
Now equating it with the obtained sum which is \[{n^2}\] .
\[25 = {n^2}\]
Taking square roots on both sides;
\[\begin{array}{l}\sqrt {25} = \sqrt {{n^2}} \\5 = n\end{array}\]
$\therefore n=5$
So the number of odd numbers is 5.
Now the expression of \[{5^2}\]in terms of the sum of consecutive odd numbers can be obtained as we have got the number of odd numbers needed to get the sum as 25.
So the required expression is
\[{5^2} = 1 + 3 + 5 + 7 + 9\]
4) \[{11^2}\]
Let’s first calculate the value of \[{11^2}\] which is equal to \[121\] .
We have to express \[121\]in terms of the sum of consecutive odd numbers starting with \[1\] .
Here, sum of consecutive odd numbers starting with \[1\] will form an arithmetic progression i.e.
\[1 + 3 + 5 + 7 + ......n\] is an arithmetic progression as the difference between any two consecutive terms is a constant which is \[2\] here and the first term is \[1\] and let \[n\] be a number of terms.
Our series is:-\[1 + 3 + 5 + 7 + ......n\]
\[\begin{array}{l}a = 1\\d = 2\end{array}\]
Now putting values of \[a\] and\[d\] in the formula \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\], we get;
\[{S_n} = \dfrac{n}{2}\left[ {2 \times 1 + \left( {n - 1} \right) \times 2} \right]\]
Simplifying it further, we get;
\[\begin{array}{l}{S_n} = \dfrac{n}{2}\left[ {2 + 2n - 2} \right]\\{S_n} = \dfrac{n}{2}\left[ {2n} \right]\\{S_n} = {n^2}\end{array}\]
As we know from the question that the value of the sum of all terms is 121.
Now equating it with the obtained sum which is \[{n^2}\] .
\[121 = {n^2}\]
Taking square roots on both sides;
\[\begin{array}{l}\sqrt {121} = \sqrt {{n^2}} \\11 = n\end{array}\]
$\therefore n=11$
So the number of odd numbers is 11 .
Now the expression of \[{11^2}\] in terms of the sum of consecutive odd numbers can be obtained as we have got the number of odd numbers needed to get the sum as 121.
So the required expression is
\[{11^2} = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21\]
Note:
Since we have used arithmetic progression here, we need to understand it properly.
An arithmetic progression is a sequence of numbers in which the difference of any two consecutive numbers of the series is a constant.
There are some properties of arithmetic progression which should be kept in mind:-
If the same number is added or subtracted from each term of an arithmetic progression, then the resulting terms in arithmetic progression are also in arithmetic progression.
If each term of an arithmetic progression is divided or multiplied with the same non-zero number, then the resulting terms in arithmetic progression are also in arithmetic progression.
Also, there is another formula for sum of arithmetic series if the first term and last term of the series is known:
\[{S_n} = \dfrac{n}{2}\left[ {{a_1} + l} \right]\]
Where \[{a_1} = \] first term
\[l = \] last term
\[n = \] number of terms.
Here we have to represent those numbers in terms of the sum of consecutive odd numbers. This will be an arithmetic series since the difference of any two consecutive terms is a constant. So we will use the formula of sum of arithmetic series here. From this formula we will get the number of terms of consecutive odd numbers and then we will express the given number in terms of the sum of consecutive odd numbers.
Complete step by step solution:
1) We know that the value of \[{7^2}\] is equal to 49.
We have to express 49 in terms of the sum of consecutive odd numbers starting with 1.
Here, sum of consecutive odd numbers starting with 1 will form an arithmetic progression i.e.
\[1 + 3 + 5 + 7 + ......n\] is an arithmetic progression as the difference between any two consecutive terms is a constant which is \[2\] here and the first term is \[1\] and let \[n\] be the number of terms.
As we know
Our series is:- \[1 + 3 + 5 + 7 + ......n\]
\[\begin{array}{l}a = 1\\d = 2\end{array}\]
Now putting values of \[a\] and\[d\] in the formula, we get;
\[{S_n} = \dfrac{n}{2}\left[ {2 \times 1 + \left( {n - 1} \right) \times 2} \right]\]
Simplifying it further, we get;
\[\begin{array}{l}{S_n} = \dfrac{n}{2}\left[ {2 + 2n - 2} \right]\\{S_n} = \dfrac{n}{2}\left[ {2n} \right]\\{S_n} = {n^2}\end{array}\]
As we know from the question that the value of sum of all terms is \[49\]
Now equating it with the obtained sum which is \[{n^2}\] .
\[49 = {n^2}\]
Taking square roots on both sides;
\[\begin{array}{l}\sqrt {49} = \sqrt {{n^2}} \\7 = n\end{array}\]
$\therefore n=7$
So the number of odd numbers is 7.
Now the expression of \[{7^2}\] in terms of the sum of consecutive odd numbers can be obtained as we have got the number of odd numbers needed to get the sum as 49.
So the required expression is
\[{7^2} = 1 + 3 + 5 + 7 + 9 + 11 + 13\]
2) \[{9^2}\]
Let’s first calculate the value of \[{9^2}\]which is equal to 81 .
We have to express \[81\] in terms of the sum of consecutive odd numbers starting with 1.
Here, sum of consecutive odd numbers starting with \[1\] will form an arithmetic progression i.e.
\[1 + 3 + 5 + 7 + ......n\] is an arithmetic progression as the difference between any two consecutive terms is a constant which is \[2\] here and the first term is \[1\] and let \[n\] be a number of terms.
Our series is:-\[1 + 3 + 5 + 7 + ......n\]
\[\begin{array}{l}a = 1\\d = 2\end{array}\]
Now putting values of \[a\] and\[d\] in the formula \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\], we get;
\[{S_n} = \dfrac{n}{2}\left[ {2 \times 1 + \left( {n - 1} \right) \times 2} \right]\]
Simplifying it further, we get;
\[\begin{array}{l}{S_n} = \dfrac{n}{2}\left[ {2 + 2n - 2} \right]\\{S_n} = \dfrac{n}{2}\left[ {2n} \right]\\{S_n} = {n^2}\end{array}\]
As we know from the question that the value of sum of all terms is \[81\]
Now equating it with the obtained sum which is \[{n^2}\] .
\[81 = {n^2}\]
Taking square roots on both sides;
\[\begin{array}{l}\sqrt {81} = \sqrt {{n^2}} \\9 = n\end{array}\]
$\therefore n=9$
So the number of odd numbers is \[9\] .
Now expression of \[{9^2}\]in terms of sum of consecutive odd numbers can be obtained as we have got the number of odd numbers needed to get the sum as \[81\]
So the required expression is
\[{9^2} = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17\]
3) \[{5^2}\]
Let’s first calculate the value of \[{5^2}\]which is equal to \[25\] .
We have to express \[25\]in terms of the sum of consecutive odd numbers starting with \[1\].
Here, sum of consecutive odd numbers starting with \[1\]will form an arithmetic progression i.e.
\[1 + 3 + 5 + 7 + ......n\] is an arithmetic progression as the difference between any two consecutive terms is a constant which is \[2\] here and the first term is \[1\] and let \[n\] be the number of terms.
Our series is:-\[1 + 3 + 5 + 7 + ......n\]
\[\begin{array}{l}a = 1\\d = 2\end{array}\]
Now putting values of \[a\] and\[d\] in the formula \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\], we get;
\[{S_n} = \dfrac{n}{2}\left[ {2 \times 1 + \left( {n - 1} \right) \times 2} \right]\]
Simplifying it further, we get;
\[\begin{array}{l}{S_n} = \dfrac{n}{2}\left[ {2 + 2n - 2} \right]\\{S_n} = \dfrac{n}{2}\left[ {2n} \right]\\{S_n} = {n^2}\end{array}\]
As we know from the question that the value of the sum of all terms is 25.
Now equating it with the obtained sum which is \[{n^2}\] .
\[25 = {n^2}\]
Taking square roots on both sides;
\[\begin{array}{l}\sqrt {25} = \sqrt {{n^2}} \\5 = n\end{array}\]
$\therefore n=5$
So the number of odd numbers is 5.
Now the expression of \[{5^2}\]in terms of the sum of consecutive odd numbers can be obtained as we have got the number of odd numbers needed to get the sum as 25.
So the required expression is
\[{5^2} = 1 + 3 + 5 + 7 + 9\]
4) \[{11^2}\]
Let’s first calculate the value of \[{11^2}\] which is equal to \[121\] .
We have to express \[121\]in terms of the sum of consecutive odd numbers starting with \[1\] .
Here, sum of consecutive odd numbers starting with \[1\] will form an arithmetic progression i.e.
\[1 + 3 + 5 + 7 + ......n\] is an arithmetic progression as the difference between any two consecutive terms is a constant which is \[2\] here and the first term is \[1\] and let \[n\] be a number of terms.
Our series is:-\[1 + 3 + 5 + 7 + ......n\]
\[\begin{array}{l}a = 1\\d = 2\end{array}\]
Now putting values of \[a\] and\[d\] in the formula \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\], we get;
\[{S_n} = \dfrac{n}{2}\left[ {2 \times 1 + \left( {n - 1} \right) \times 2} \right]\]
Simplifying it further, we get;
\[\begin{array}{l}{S_n} = \dfrac{n}{2}\left[ {2 + 2n - 2} \right]\\{S_n} = \dfrac{n}{2}\left[ {2n} \right]\\{S_n} = {n^2}\end{array}\]
As we know from the question that the value of the sum of all terms is 121.
Now equating it with the obtained sum which is \[{n^2}\] .
\[121 = {n^2}\]
Taking square roots on both sides;
\[\begin{array}{l}\sqrt {121} = \sqrt {{n^2}} \\11 = n\end{array}\]
$\therefore n=11$
So the number of odd numbers is 11 .
Now the expression of \[{11^2}\] in terms of the sum of consecutive odd numbers can be obtained as we have got the number of odd numbers needed to get the sum as 121.
So the required expression is
\[{11^2} = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21\]
Note:
Since we have used arithmetic progression here, we need to understand it properly.
An arithmetic progression is a sequence of numbers in which the difference of any two consecutive numbers of the series is a constant.
There are some properties of arithmetic progression which should be kept in mind:-
If the same number is added or subtracted from each term of an arithmetic progression, then the resulting terms in arithmetic progression are also in arithmetic progression.
If each term of an arithmetic progression is divided or multiplied with the same non-zero number, then the resulting terms in arithmetic progression are also in arithmetic progression.
Also, there is another formula for sum of arithmetic series if the first term and last term of the series is known:
\[{S_n} = \dfrac{n}{2}\left[ {{a_1} + l} \right]\]
Where \[{a_1} = \] first term
\[l = \] last term
\[n = \] number of terms.
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