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Mathematical Induction — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSMathematical Induction4 Marks
Question
Prove the following by the principle of mathematical induction: $1.3 + 3.5 + 5.7 + ... + (2\text{n} - 1)(2\text{n} + 1)= \frac{\text{n}(4\text{n}^2+6\text{n}-1)}{3}$
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Answer

Let P(n): $1.3 + 3.5 + 5.7 + ... +(2\text{n} - 1)(2\text{n} + 1)= \frac{\text{n}(4\text{n}^2+6\text{n}-1)}{3}$ for n = 1 $1.3=\frac{1(4+6-1)}{3}$ 3 = 3 ⇒ P(n) is true for n = 1 Let P(n) is true for n = k, so $1.3 + 3.5 + 5.7 + ... +(2\text{k} - 1)(2\text{k} + 1)= \frac{\text{k}(4\text{k}^2+6\text{k}-1)}{3} \ ...(1)$ We have to show that $1.3 + 3.5 + 5.7 + ... +(2\text{k} - 1)(2\text{k} + 1)+(2\text{k} + 1)(2\text{k} + 3) $ $=\frac{(\text{k}+1)\big[4(\text{k} + 1)^2+6(\text{k+1})-1\big]}{3}$ Now, ${1.3 + 3.5 + 5.7 + ... + (2\text{k} - 2)(2\text{k} + 1)} + (2\text{k} + 1)(2\text{k} + 3)$ $=\frac{\text{k}(4\text{k}^2+6\text{k}-1)}{3}+(2\text{k}+1)(2\text{k}+3)$ [Using equation (1)] $=\frac{\text{k}(4\text{k}^2+6\text{k}-1)+3(4\text{k}^2+6\text{k}+2\text{k}+3)}{3}$ $=\frac{4\text{k}^3+6\text{k}-​\text{k}​+12\text{k}^2+18\text{k}+6\text{k}+9}{3}$ $=\frac{4\text{k}^3+18\text{k}^2+23\text{k}+9}{3}$ $=\frac{4\text{k}^3+4\text{k}^2​+14\text{k}^2+14\text{k}+9\text{k}+9}{3}$ $=\frac{(\text{k}+1)(4\text{k}^2+8\text{k}+4+6\text{k}+6-1)}{3}$ $=\frac{(\text{k}+1)\big[(4(\text{k}+1)^2+6(\text{k}+1)-1\big]}{3}$ ⇒ P(n) is true for n = k + 1 ⇒ P(n) is true for all $\text{n}\in\text{N}$ by PMI

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