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Mathematical Induction — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsMathematical Induction5 Marks
Question
Prove the following by the principle of mathematical induction:$\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{n}-1)\text{d})=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
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Answer

Let p(n): $\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{n}-1)\text{d})=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For n = 1
$\text{a}=\frac{1}{2}[2\text{a}+(1-1)\text{d}]$
⇒ p(n) is true for n = 1
Let p(n) is true for n = k, so
$\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{k}-1)\text{d})=\frac{\text{k}}{2}[2\text{a}+(\text{k}-1)\text{d}] \ ...(1)$
We have to show that
$\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{k}-1)\text{d})+(\text{a}+(\text{k})\text{d})=\frac{(\text{k+1})}{2}[2\text{a}+\text{k}\text{d}] $
Now,
$\big\{\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{k}-1)\text{d})\big\}+(\text{a}+\text{k}\text{d})$
$\frac{\text{k}}{2}[2\text{a}+(\text{k}-1)\text{d}]+(\text{a}+\text{k}\text{d})$ [Using equation (1)]
$=\frac{2\text{k}\text{a}+\text{k}(\text{k}-1)\text{d}+2(\text{a}+\text{k}\text{d})}{2}$
$=\frac{2\text{k}\text{a}+\text{k}^2\text{d}-\text{k}\text{d}+2\text{a}+2\text{k}\text{d})}{2}$
$=\frac{2\text{k}\text{a}+2\text{a}+\text{k}^2\text{d}+\text{k}\text{d}}{2}$
$=\frac{2\text{a}(\text{k}+1)+\text{d}(\text{k}^2 +\text{k})}{2}$
$=\frac{(\text{k}+1)}{2}[2\text{a}+\text{k}\text{d}]$
⇒ p(n) is true for n = k + 1
⇒ p(n) is true for all by $\text{n}\in\text{N}$ PMI

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