<p>Observe the following: 1<sup>2</sup> = (1*2*3)/6<br />   
ID: 2944576 • Letter: #
Question
<p>Observe the following: 1<sup>2</sup> = (1*2*3)/6<br />                                1<sup>2</sup> + 3<sup>2</sup> = (3*4*5)/6<br />                         1<sup>2</sup> + 3<sup>2</sup> + 5<sup>2</sup> = (5*6*7)/6<br />Guess a general law suggested by these examples and provide an algebraic proof of why it works. This may help in your proof: for any integer k is greater than or equal to 1, the following equality holds:</p><p>1<sup>2</sup> + 2<sup>2</sup> + 3<sup>2</sup> + 4<sup>2 </sup>+...+k<sup>2</sup> = (k(k+1)(2k+1))/6</p>
Explanation / Answer
simple..use mathematical induction to show 1^2 + 2^2 + 3^2 + 4^2 +...+k^2 = (k(k+1)(2k+1))/6 step 1: for k=1 1^2 = (1*2*3)/6 ; this holds, since this is given in the question Step 2: let for k=n, this relation is true i.e 1^2 + 2^2 + 3^2 + 4^2 +...+n^2 = (n(n+1)(2n+1))/6 is true step 3: now, we have to show that for k=n+1, it is also true. 1^2 + 2^2 + 3^2 + 4^2 +...+n^2 +(n+1)^2= ((n+1)(n+2)(2(n+1)+1))/6 =>(n(n+1)(2n+1))/6 + (n+1)^2 =((n+1)(n+2)(2n+3))/6 =>(n+1)[ n(2n+1)/6 + (n+1)] =(n+1) (n+2)(2n+3)/6 =>(n+1)/6[ n(2n+1) + 6(n+1)] = (n+1) (n+2)(2n+3)/6 =>(n+1)/6[ 2n^2+n + 6n+6)] = (n+1) (n+2)(2n+3)/6 =>(n+1)/6[ 2n^2+7n+6)] = (n+1) (n+2)(2n+3)/6 =>(n+1)/6[ 2n^2+4n+3n+6)] = (n+1) (n+2)(2n+3)/6 =>(n+1)/6[ 2n(n+2)+3(n+2)] = (n+1) (n+2)(2n+3)/6 =>(n+1)(2n+3)(n+2)/6 = (n+1) (n+2)(2n+3)/6 LHS=RHS hence by mathematics induction, the relation 1^2 + 2^2 + 3^2 + 4^2 +...+k^2 = (k(k+1)(2k+1))/6 is proved
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