CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Prove by the principle of mathematical induction that 1+2+3+...+n<(2n+1)28, for all nN.

         Or

Using the principle of mathematical induction, prove that

1.3+2.32+3.33+...+n.3n=(2n1)3n+1+34. for all nN.


Solution

Let P(n) be the statement given by

P(n):1+2+3+...+n<(2n+1)28

For n=1, P(1):1<(2+1)28

           1<98

   P(1) is true.

Assume that, P(k) be true, for some natural number k.

1+2+3+...+k <(2k+1)28

Now, we will show that P(k+1) is true whenever P(k) is true.

i.e. 1+2+3+...+k+k+1<{2(k+1)+1}28

Now, P(k) is true.

           1+2+3+...+k <(2k+1)28

           1+2+3+...+k+(k+1)<(2k+1)28+(k+1)

                                                                       [adding both sides by (k+1)]

           1+2+3+...+k+(k+1)<18{(2k+1)2+8(k+1)}

           1+2+3+...+k+(k+1)<18(4k2+4k+1+8k+8)

           1+2+3+...+k+(k+1)<18(4K2+12K+9)

           1+2+3+...+K+(K+1)<18(2k+3)2

           1+2+3+...+k+(k+1)<18{2(k+1)+1}2

Thus, P(k+1) is true.

Hence, by principle of mathematical induction, P(n) is true,  nN.

Let P(n) be the statement given by

P(n):1.3+2.32+3.32+...+n.3n=(2n1)3n+1+34

For          n=1:P(1):l.3={2(1)1}31+1+34

           3=9+34  3=3

So, P(1) is true.

Assume that P(k) be true, for some natural number k.

 1.3+2.32+3.33+...+k.3k=(2k1)3k+1+34        ...(i)

Now, we will show that P(k+1) is true, whenever P(k) is true

i.e. 1.3+2.32+3.33 +...+k.3k+(k+1).3k+1

= {2(k+1)1}3(k+1)+1+34

Now, 1.3+2.32+3.32 +...+k.3k+(k+1).3k+1

= (2k1)3k+1+34+(k+1).3k+1         [from Eq.  (i)]

= (2k1)3k+1+3+4(k+1).3k+14

= (2k1+4k+4).3k+1+34

= (6k+3).3k+1+34

= (2k+1).3k+2+34

= {2(k+1)1}.3(k+1)+1+34

  P(k+1) is true.

Hence, by principle of mathematical induction, P(n) is true,  nN.

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image