Question

1^2=1
1^2+2^2=5
1^2+2^2+3^2=14
1^2+2^2+3^2+4^2=30
1^2+2^2+3^2+4^2+....+n^2=an^3+bn^2+n/6. work out the values of a and b

Answer 1

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

• The sum of squares of the first n natural numbers is :- [ n(n + 1)(2n + 1) ] / 6

Sᴏʟᴜᴛɪᴏɴ :-

→ [ n(n + 1)(2n + 1) ] / 6

Lets Solve Numerator Part,

→ n(n + 1)(2n + 1)

→ (n² + n)(2n + 1)

→ n²(2n + 1) + n(2n + 1)

→ 2n³ + n² + 2n² + n

→ 2n³ + 3n² + n

So ,

→ 1² +2² +3² +4² +....+n² = (2n³ + 3n² + n) / 6

Comparing now , we get,

→ (2n³ + 3n² + n) / 6 = ( an³+bn²+n ) / 6

Hence,

→ a = 2 (Ans.)

→ b = 3 (Ans.)

_______________

Check The formula :-

→ 1² = [ 1 * 2 * 3 ] / 6 = 6/6 = 1

→ 1² + 2² = (2 * 3 * 5) / 6 = 5

→ 1² + 2² + 3² = (3 * 4 * 7) / 6 = 14

→ 1² + 2² + 3² + 4² = (4 * 5 * 9) / 6 = 30

So,

→ 1² + 2² + 3² + 4² + ______ n² = (2n³ + 3n² + n) / 6 .

Answer 2

GIVEN SEQUENCE :-

1² = 1

1² + 2² = 5

1² + 2² + 3² = 14

1² + 2² + 3² + 4² = 30

TO OPERATE :-

1² + 2² + 3² + 4² + ..... + n² = an³ + bn2 + n/6

TO FIND THE VALUE OF :-

a and b.

SOLUTION :-

We know,

✰ sum of first n natural numbers is :

$\displaystyle{\sf{\frac{n(n+1)}{2} }}$

✰ And, the sum of the squares of the first n natural numbers is :

$\displaystyle{\sf{\frac{n(n+1)(2n+1)}{6} }}$

We are required to find the sum of squares of first n natural numbers. So, on solving, we get :

$\displaystyle{\sf{\frac{(n^2+n)(2n+1)}{6} }}\\\\\\\displaystyle{=\sf{\frac{2n^3+n^2+2n^2+n}{6} }}\\\\\\\displaystyle{=\sf{\frac{2n^3+3n^2+n}{6} }}$

$\textsf{Comparing it with}\ \displaystyle{\sf{\frac{an^3+bn^2+n}{6} }},\ \sf{we\ get\ :}$

$\Large\underline{\boxed{\boxed{a = 2}}}$

$\Large\underline{\boxed{\boxed{b = 3}}}$

$\rule{200}{3}$