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

Hint: 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.