Question

Sum of the squares of n natural numbers exceeds their sum by $330$ , then n equals?
1. $8$
2. $10$
3. $15$
4. $20$

Answer 1

Before solving this question one must know that natural numbers are the whole numbers except $0$. $0$ is not included in natural numbers meaning that natural numbers start from $1,2,3..$ . Now to solve this question we must know two formulas. First the formula for sum of n numbers in natural numbers. The formula to find the sum of n natural numbers is $\dfrac{n(n+1)}{1}$
Now we must also know the formula to find the sum of squares of first n natural numbers. Therefore the formula to find the sum of squares of first n natural numbers is $\dfrac{n(n+1)(2n+1)}{6}$

Complete step-by-step solution:
Here in this question we are asked to find the n when the sum of squares of first n natural numbers exceeds the sum of n natural numbers by $330$ . Now here we need to find the n.
For that we must know what the formula to find the sum of first n natural numbers is. The formula is
Sum of squares of first n natural numbers $=\dfrac{n(n+1)(2n+1)}{6}$
Now as said in the question, the sum of the squares of the first n natural numbers exceeds the sum of those n natural numbers. Hence we must also know the formula to find the sum of first n natural numbers
Sum of first n natural numbers $=\dfrac{n(n+1)}{2}$
Now according to the question we can make the equation that
$\dfrac{n(n+1)(2n+1)}{6}-\dfrac{n(n+1)}{2}=330$
Taking the LCM and cross multiplying we get
$n(n+1)(2n+1)-3n(n+1)=1980$
Now taking similar sum out we get
$n(n+1)[2n+1-(3)]=1980$
$n(n+1)[2n-2]=1980$
Dividing both sides by $2$ we get
$n(n+1)(n-1)=990$
Finding factors of $990$ we get
$n(n+1)(n-1)=10\times 11\times 9$
Hence we found the value of n which is $n=10$.

Note: We must know that natural numbers are part of the number system. These numbers are positive integers which start from $1$ and exist till infinity. Hence we can say that there are infinite natural numbers.