Question

The sum of the squares of the first n positive integers is given by the expression $\frac{n(n+c)(2n+k)}{6}$, if c and k are, respectively:

• A. 1 and 2
• B. 3 and 5
• C. 2 and 2
• D. 1 and 1

Answer 1

The correct option is D 1 and 1

Given: $S=1^2+2^2+3^2.......n^2=\frac{n(n+c)(2n+k)}{6}$

find: Value of $C$ and $K$

Solution: We know that

$(x+1{)}^3=4^3+{3.4}^2+3.x+1$

Now,

$(1+1{)}^3=1^3+{3.1}^2+3.1+1$

$(2+1{)}^3=2^3+{3.2}^2+3.2+1$

$(3+1{)}^3=3^3+{3.3}^2+3.3+1$

$(n+1{)}^3=n^3+3.n^2+3.n+1$

When we add all of these volume then we get

$2^3+3^3+4^3......(n+1{)}^3=1^3+2^3+3^3......n^3+3(1^2+2^2+.......n^2)+3(1+\mathrm{2.....}n)+n$

$(n+1{)}^3=1^3+3(5)+3(1+2+\mathrm{3.....}n)+n$

We Know that $1+2+\mathrm{3....}n=\frac{n(n+1)}{2}$

$(n+1{)}^3+1+35+3n\frac{(n+1)}{2}+n$

$(n+1{)}^3-3n\frac{(n+1)}{2}-(n+1)=35$

$(n+1)\left[\right(n+1{)}^2-\frac{3n}{2}-1]=35$

$(n+1)[n^2+2n+1-\frac{3n}{2}-1]=35$

$(n+1)\left[\frac{2n^2+4n-3n}{2}\right]=35$

$(n+1)\left(\frac{2n^2+x}{6}\right)=5$

$n(n+1)\frac{(2n+1)}{6}=5$

Compare both value and get $C=1$ , $k=1$