![Ex 4.1, 11 - Chapter 4 Class 11 Mathematical Induction - Part 2](https://d1avenlh0i1xmr.cloudfront.net/2789c442-aaf6-4798-9984-d6022bbc50c4slide10.jpg)
![Ex 4.1, 11 - Chapter 4 Class 11 Mathematical Induction - Part 3](https://d1avenlh0i1xmr.cloudfront.net/a4837fe0-8b2c-4afa-86bc-4b450e31f84dslide11.jpg)
![Ex 4.1, 11 - Chapter 4 Class 11 Mathematical Induction - Part 4](https://d1avenlh0i1xmr.cloudfront.net/77f6dc2a-2bd0-4786-8fb2-1dd33ef8eb43/slide8.jpg)
![Ex 4.1, 11 - Chapter 4 Class 11 Mathematical Induction - Part 5](https://d1avenlh0i1xmr.cloudfront.net/c5a811fc-1e9c-4d82-ae3b-2bbdacdd73bd/slide9.jpg)
![Ex 4.1, 11 - Chapter 4 Class 11 Mathematical Induction - Part 6](https://d1avenlh0i1xmr.cloudfront.net/dbaaaa75-2f7a-4dfb-bdfa-6e64c4b7ddcc/slide10.jpg)
Mathematical Induction - Questions and Solutions
Last updated at April 16, 2024 by Teachoo
Question11 Prove the following by using the principle of mathematical induction for all n ∈ N: 11.2.3 + 12.3.4 + 13.4.5 + ……..+ 1𝑛 𝑛 + 1(𝑛 + 2) = 𝑛(𝑛 + 3)4(𝑛 + 1)(𝑛 + 2) Let P (n) : 11.2.3 + 12.3.4 + 13.4.5 + ……..+ 1𝑛 𝑛 + 1(𝑛 + 2) = 𝑛(𝑛 + 3)4(𝑛 + 1)(𝑛 + 2) For n = 1, L.H.S = 11.2.3 = 16 R.H.S = 1.(1 + 3)4(1 + 1)(1 + 2) = 1.44.2.3 = 12.3 = 16 Hence, L.H.S. = R.H.S , ∴ P(n) is true for n = 1 Assume P(k) is true 11.2.3 + 12.3.4 + 13.4.5 + ……..+ 1𝑘 𝑘 + 1(𝑘 + 2) = 𝑘(𝑘 + 3)4(𝑘 + 1)(𝑘 + 2) We will prove that P(k + 1) is true. R.H.S = 𝑘 + 1 𝑘 + 1+ 34 𝑘 + 1+ 1 𝑘 + 1+ 2 L.H.S = 11.2.3 + 12.3.4 + 13.4.5 + ……..+ 1 𝑘 + 1 𝑘 + 1+ 1( 𝑘 + 1+ 2) Rough Factorizing 𝐤𝟑+ 𝟒+ 𝟗𝐤+ 𝟔𝐤𝟐 Let f(k) = k3+ 4+ 9k+ 6k2 Putting k = − 1 f(−1) = (−1)3+ 4+9 −1+6 (−1)2 = −1 + 4− 9+6 = 0 Thus, (k + 1) is a factor of f(k) ∴ f(k) = (𝑘+1)( 𝑘2+5𝑘+4)