NCERT Solutions for Class 11 Mathematics Chapter 1 Sets

Sets Here is the CBSE Mathematics Chapter 1 for Class 11 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 11 Mathematics Sets Chapter 1 NCERT Solutions for Class 11 Mathematics Sets Chapter 1 The following is a summary in Hindi and English for the academic year 2025-26. You can save these solutions to your computer or use the Class 11 Mathematics.

Question 1

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Use principle of mathematical induction to prove that:

$1 space plus space 2 space plus space 3 space plus space... space plus space straight n space equals space fraction numerator straight n left parenthesis straight n space plus space 1 right parenthesis over denominator 2 end fraction$

Solution

Let P(n): 1 + 2 + 3 + ......... + n = $space space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction$
I. For n = 1,
    P(1) : 1 = $fraction numerator 1 left parenthesis 1 plus 1 right parenthesis over denominator 2 end fraction rightwards double arrow space 1 space equals space 1 space rightwards double arrow space space straight P left parenthesis 1 right parenthesis$ is true.
II. Suppose the statement is true for n = m, $straight m element of straight N$
      i.e. P(m): $1 plus 2 plus 3 plus........ space plus straight m space equals space fraction numerator straight m left parenthesis straight m plus 1 right parenthesis over denominator 2 end fraction$           ....(i)
III.    For n = m + 1,
        P(m + 1): 1 + 2 + 3 + ........ + (m + 1) = $fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction$
or [1 + 2 + 3 + ...... + m] + (m + 1) = $fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction$

         [From (i), 1 + 2 + 3 + ...... + m = $fraction numerator straight m left parenthesis straight m plus 1 right parenthesis over denominator 2 end fraction$ ]
∴        P (m + 1): $space fraction numerator straight m left parenthesis straight m space plus space 1 right parenthesis over denominator 2 end fraction space plus space left parenthesis straight m space plus space 1 right parenthesis space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction$
$rightwards double arrow$ $space space left parenthesis straight m plus 1 right parenthesis open parentheses straight m over 2 plus 1 close parentheses space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction$
$rightwards double arrow$    $left parenthesis straight m plus 1 right parenthesis open parentheses fraction numerator straight m plus 2 over denominator 2 end fraction close parentheses space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction$
$rightwards double arrow$ $fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction$
    which is true

∴ P(m + 1) is true

∴ P(m) is true $rightwards double arrow$ P(m + 1) is true
Hence, by mathematical induction
P(n) is true for all $space space straight n element of straight N.$