Question

1 square +2 square +3 square +4 square............+n square = n(n+1)(2n+1) whole divided by 6

Answer 1

Answer:

Proof below:

Step-by-step explanation:

1² + 2² + 3² + 4² + . . . . . . . . . . . . . + n² =  $\frac{n(n+1)(2n+1)}{6}$

We use Mathematical Induction to prove the following

For n = 1

LHS = 1² = 1

RHS =  $\frac{1(1+1)(2*1+1)}{6}$ =  $\frac{1(2)(3)}{6}$ =    $\frac{6}{6}$ = 1

As LHS = RHS, it is true for n = 1

For n = k

It it hold true for n = k

=> 1² + 2² + 3² + 4² + . . . . . . . . . . . . . + k² =  $\frac{k(k+1)(2k+1)}{6}$

For n = k+1

For this, we need to show that

1² + 2² + 3² + 4² + . . . . . . . . . . . . . + (k+1)² =  $\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$

1² + 2² + 3² + 4² + . . . . . . . . . . . . . + (k+1)² =  $\frac{(k+1)(k+2)(2k+3)}{6}$

Proof:

Taking LHS

= 1² + 2² + 3² + 4² + . . . . . . . . . . . . . + (k+1)²

=1² + 2² + 3² + 4² + . . . . . . . . . . . . . k² + (k+1)²

As we have already proved that

1² + 2² + 3² + 4² + . . . . . . . . . . . . . + k² =  $\frac{k(k+1)(2k+1)}{6}$

Therefore, on substituting, we get,

= $\frac{k(k+1)(2k+1)}{6}$ + (k+1)²

= $\frac{k(k+1)(2k+1) + 6(k+1)^{2}}{6}$        (Taking LCM)

= $\frac{(k+1)[2(2k+1) + 6(k+1)]}{6}$        (Taking common)

= $\frac{(k+1)(2k^{2}+k + 6k+6)}{6}$           (Opening the brackets)

= $\frac{(k+1)(2k^{2}+7k+6)}{6}$  

= $\frac{(k+1)(2k^{2}+4k+3k +6)}{6}$

=  $\frac{(k+1)(2k(k+2)+3(k+2)}{6}$

=  $\frac{(k+1)((k+2)(2k+3)}{6}$

= RHS

Therefore, it is true for n = k+1

As it is true for n = k+1, it will be true for all values of n

Hence proved.    

Answer 2

Step-by-step explanation:

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