Dear Student , Please find below the solution to the asked query : Let Pn be the given statement,Pn: 12+22+ +n2 > n33 , n∈NFor n=112>1331>13It is true Let us assume that it is also true for n=k 12+22+ +k2 > k33Now,For n=k+1 12+22+ +k2 +k+12>k33+k+12 Since 12+22+ +k2 > k33 =k3+3k2+6k+33=k3+3k2+3k+1+3k+23=k3+3k21+3k12+12+3k+23=k+13+3k+23>k+133So , it is also true for n=k+1 Hence, By PMI Pn is true for all n∈N Hope this information will clear your doubts about the topic If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible Regards

prove by pmi

prove by pmi PRINCIPLE 'OF MATæ e 7 Pro e that 3 n 12 +22 + > 3

Dear Student ,

Please find below the solution to the asked query :

Let P (n) be the given statement, P (n) : 1^{2} + 2^{2} + . . . + n^{2} > \frac{n^{3}}{3}, n \in N For n = 1 1^{2} > \frac{1^{3}}{3} 1 > \frac{1}{3} It is true . Let us assume that it is also true for n = k 1^{2} + 2^{2} + . . . + k^{2} > \frac{k^{3}}{3} Now, For n = k + 1 1^{2} + 2^{2} + . . . + k^{2} + {(k + 1)}^{2} > \frac{k^{3}}{3} + {(k + 1)}^{2} (Since 1^{2} + 2^{2} + . . . + k^{2} > \frac{k^{3}}{3}) = \frac{k^{3} + 3 k^{2} + 6 k + 3}{3} = \frac{k^{3} + 3 k^{2} + 3 k + 1 + 3 k + 2}{3} = \frac{{(k)}^{3} + 3 {(k)}^{2} (1) + 3 (k) {(1)}^{2} + {(1)}^{2} + 3 k + 2}{3} = \frac{{(k + 1)}^{3} + 3 k + 2}{3} > \frac{{(k + 1)}^{3}}{3} So, it is also true for n = k + 1 . Hence, By PMI P (n) is true for all n \in N .

Hope this information will clear your doubts about the topic .

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible .

Regards

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prove by pmi prove by pmi PRINCIPLE 'OF MATæ e 7 Pro e that 3 n 12 +22 + > 3

prove by pmi

prove by pmi PRINCIPLE 'OF MATæ e 7 Pro e that 3 n 12 +22 + > 3