Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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2 + 5 + 8 + 11 + ... + (3n − 1) = 1 2 N ( 3 N + 1 ) - Mathematics

2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]

 
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Solution

Let P(n) be the given statement.
Now,

\[P(n) = 2 + 5 + 8 + . . . + (3n - 1) = \frac{1}{2}n(3n + 1)\]

\[\text{ Step} 1: \]

\[P(1) = 2 = \frac{1}{2} \times 1(3 + 1) \]

\[\text{ Hence, P(1) is true .}  \]

\[\text{ Step}  2: \]

\[\text{ Let P(m) be true } . \]

\[\text{ Then, } \]

\[2 + 5 + 8 + . . . + (3m - 1) = \frac{1}{2}m(3m + 1)\]

\[\text{ To prove: P(m + 1) is true } . \]

\[\text{ That is, } \]

\[2 + 5 + 8 + . . . + (3m + 2) = \frac{1}{2}(m + 1)(3m + 4)\]

\[\text{ P(m) is equal to:}  \]

\[2 + 5 + 8 + . . . + (3m - 1) = \frac{1}{2}m(3m + 1)\]

\[\text{ Thus, we have:}  \]

\[2 + 5 + 8 + . . . + (3m - 1) + (3m + 2) = \frac{1}{2}m(3m + 1) + (3m + 2) \left[ \text{ Adding } (3m + 2) \text{ to both sides } ] \right]\]

\[ \Rightarrow 2 + 5 + 8 + . . . + (3m + 2) = \frac{1}{2}(3 m^2 + m + 6m + 4) = \frac{1}{2}(3 m^2 + 7m + 4)\]

\[ \Rightarrow 2 + 5 + 8 + . . . + (3m + 2) = \frac{1}{2}(3m + 4)(m + 1)\]

\[\text{ Thus, P(m + 1) is true } . \]

\[\text{ By the principle of mathematical induction, P(n) is true for all n}  \in N .\]

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 11 | Page 27
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