Find the sum of all natural numbers from 11 to 30, using the expression
[tex] \frac{1}{2} {n}^{2} + \frac{1}{2}n[/tex]
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Find the sum of all natural numbers from 11 to 30, using the expression
[tex] \frac{1}{2} {n}^{2} + \frac{1}{2}n[/tex]
pls answer as soon as possible
wrong answers will be reported
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[tex]\large\underline{\sf{Solution-}}[/tex]
We have to find the sum of all natural numbers from 11 to 30.
That means,
[tex]\sf \: 11 + 12 + 13 + ... + 30 \\ \\ [/tex]
can be rewritten as
[tex]\sf \: = \: (1 + 2 + 3 + ... + 30) - (1 +2 + 3 + .. + 10) \\ \\ [/tex]
Now, it is given that
[tex]\boxed{ \sf{1 + 2 + 3 + ... + n = \dfrac{1}{2} {n}^{2} + \dfrac{1}{2}n \: }} \\ \\ [/tex]
So, using this result, we get
[tex]\sf \: = \: \dfrac{1}{2} {(30)}^{2} + \dfrac{1}{2}(30) - \bigg(\dfrac{1}{2} {(10)}^{2} + \dfrac{1}{2}(10) \bigg) \\ \\ [/tex]
[tex]\sf \: = \: \dfrac{1}{2} (900) + 15 - \bigg(\dfrac{1}{2} (100) + 5\bigg) \\ \\ [/tex]
[tex]\sf \: = \: 450 + 15 - 50 - 5 \\ \\ [/tex]
[tex]\sf \: = \: 465 - 55 \\ \\ [/tex]
[tex]\sf \: = \: 410 \\ \\ [/tex]
Hence,
[tex]\bf\implies \: 11 + 12 + 13 + ... + 30 = 410\\ \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]
Step-by-step explanation:
[tex] \pmb{ \bf{Given \: that, }}[/tex]
The sum of first n natural numbers is
[tex] \sf \: \frac{1}{2} \: {n}^{2} + \frac{1}{2}n \\ [/tex]
[tex] \pmb{ \bf{To \: find }}[/tex]
The sum of natural numbers from 11 to 30.
We know that,
Sum of natural numbers from 11 to 30
= sum of first 30 natural numbers - sum of first 10 natural numbers.
Sum of first 30 natural numbers, (n=30)
[tex]\sf\frac{1}{2} n^2 + \frac{1}{2} n = \frac{1}{2} (30)^2+ \frac{1}{2} (30) \\ \\ \\ \sf=\frac{1}{2} (900) + \frac{1}{2} (30) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf= \frac{1}{2} (900 + 30)\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf= \frac{1}{2} (930)\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \red{ \bf= 465}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ [/tex]
Sum of first 10 natural numbers,( n = 10 )
[tex] \sf \frac{1}{2} n ^{2} + \frac{1}{2}n = \frac{1}{2}(10)^{2} +\frac{1}{2} (10) \\ \\ \\ \sf= \frac{1}{2} (100) +\frac{1}{2} (10) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf= \frac{1}{2} (100 + 10) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf =\frac{1}{2}(110) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \red{ \bf= 55}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ [/tex]
Sum of natural numbers from 11 to 30
[tex] \sf \: = 465 - 55 \\ \red{ \bf{= 410}} \: \: \: \: \: \: \: \: \\ [/tex]
Therefore,
The sum of the natural numbers from 11 to 30 is 410.
★ Additional Information :-