Question

Observe the following pattern
$$1^2=\dfrac {1}{6} [1\times (1+1)\times (2\times 1+1)]$$
$$1^2+2^2=\dfrac {1}{6} [2\times (2+1)\times (2\times 2+1)]$$
$$1^2+2^2+3^2=\dfrac {1}{6} [3\times (3+1)\times (2\times 3+1)]$$
$$1^2+2^2+3^2+4^2=\dfrac {1}{6} [4\times (4+1)\times (2\times 4+1)]$$
$$1^2+2^2+3^2+4^2+.....+10^2=?$$

Answer 1

Correct option is A. 385
Th first equality, whose biggest number on the L.H.S is $$1,$$ has $$1,1,1,2,1$$ and $$1$$ as the six numbers.

The second equality, whose biggest number on the L.H.S. is $$2,$$ has $$2,2,1,2,2$$ and $$1$$ as the six numbers.

The third equality, whose biggest number on the L.H.S is $$3,$$ has $$3,3,1,2,3$$ and $$1$$ as the six numbers.

The fourth equality, whose biggest number on the L.H.S. is $$4$$ has numbers $$4,4,1,2,4$$ and $$1$$ as the six numbers.

Note that the fourth number on R.H.S. is always $$2$$ and sixth number is always $$1.$$ The remaining numbers are equal to the biggest number on L.H.S.

Hence, if the biggest number on the L.H.S is $$n, $$ the six numbers on the R.H.S. would be $$n,n,1,2,n$$ and $$1$$.

$$1^2+2^2+3^2+.....+10^2=\dfrac{1}{6}\times [10\times (10+1)\times (2\times 10+1)]$$

$$=\dfrac{1}{6}\times [10\times 11\times 12]$$

$$=385$$