The sum of squares of first ten natural numbers is
A. 375
B. 385
C. 475
D. 485
Answer
622.2k+ views
Hint: Natural numbers are the whole numbers except 0 which means they start from 1 till infinity. First ten natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. We can find the sum of squares of these ten natural numbers using the below mentioned formula by putting n as 10.
Formula used:
Sum of squares of first ‘n’ natural numbers is $ \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} $
Complete step-by-step answer:
We are given to find the sum of squares of the first ten natural numbers.
Natural numbers are part of the number system which are the positive integers starting from 1 till infinity. There are infinite natural numbers.
The first 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
Sum of squares of first ‘n’ natural numbers is
$\Rightarrow \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} $ . Here the value of n is 10.
Therefore, their sum of squares is $ \dfrac{{10\left( {10 + 1} \right)\left( {2 \times 10 + 1} \right)}}{6} = \dfrac{{10 \times 11 \times 21}}{6} = \dfrac{{2310}}{6} = 385 $
Sum of squares of the first 10 natural numbers is 385. Hence, the correct option is Option B.
So, the correct answer is “Option B”.
Note: Another approach.
The first 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
The square of 1 is $ {1^2} = 1 $ , the square of 2 is $ {2^2} = 4 $ , the square of 3 is $ {3^2} = 9 $ , the square of 4 is $ {4^2} = 16 $ , the square of 5 is $ {5^2} = 25 $ , the square of 6 is $ {6^2} = 36 $ , the square of 7 is $ {7^2} = 49 $ , the square of 8 is $ {8^2} = 64 $ , the square of 9 is $ {9^2} = 81 $ and the square of 10 is $ {10^2} = 100 $
The sum of squares of first 10 natural numbers will be $ 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385 $
Formula used:
Sum of squares of first ‘n’ natural numbers is $ \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} $
Complete step-by-step answer:
We are given to find the sum of squares of the first ten natural numbers.
Natural numbers are part of the number system which are the positive integers starting from 1 till infinity. There are infinite natural numbers.
The first 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
Sum of squares of first ‘n’ natural numbers is
$\Rightarrow \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} $ . Here the value of n is 10.
Therefore, their sum of squares is $ \dfrac{{10\left( {10 + 1} \right)\left( {2 \times 10 + 1} \right)}}{6} = \dfrac{{10 \times 11 \times 21}}{6} = \dfrac{{2310}}{6} = 385 $
Sum of squares of the first 10 natural numbers is 385. Hence, the correct option is Option B.
So, the correct answer is “Option B”.
Note: Another approach.
The first 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.
The square of 1 is $ {1^2} = 1 $ , the square of 2 is $ {2^2} = 4 $ , the square of 3 is $ {3^2} = 9 $ , the square of 4 is $ {4^2} = 16 $ , the square of 5 is $ {5^2} = 25 $ , the square of 6 is $ {6^2} = 36 $ , the square of 7 is $ {7^2} = 49 $ , the square of 8 is $ {8^2} = 64 $ , the square of 9 is $ {9^2} = 81 $ and the square of 10 is $ {10^2} = 100 $
The sum of squares of first 10 natural numbers will be $ 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385 $
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