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Home » Solve: 1.2 + 2.3 + 3.4 + … + n(n+1) = [n (n+1) (n+2)] / 3
Solve: 1.2 + 2.3 + 3.4 + … + n(n+1) = [n (n+1) (n+2)] / 3
CBSE | Class 11 | Exercise 12.2 | Mathematical Induction | Maths | RD Sharma
Solve: 1.2 + 2.3 + 3.4 + … + n(n+1) = [n (n+1) (n+2)] / 3

Let ,

P(n): given equation

Let us check for

    \[n\text{ }=\text{ }1,\]

    \[P\text{ }\left( 1 \right):\text{ }1\left( 1+1 \right)\text{ }=\text{ }\left[ 1\left( 1+1 \right)\text{ }\left( 1+2 \right) \right]\text{ }/3\]

    \[:\text{ }2\text{ }=\text{ }2\]

P (n) is true for

    \[n\text{ }=\text{ }1.\]

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1)

is true.

    \[P\text{ }\left( k \right):\text{ }1.2\text{ }+\text{ }2.3\text{ }+\text{ }3.4\text{ }+\text{ }\ldots \text{ }+\text{ }k\left( k+1 \right)\text{ }=\text{ }\left[ k\text{ }\left( k+1 \right)\text{ }\left( k+2 \right) \right]\text{ }/\text{ }3\text{ }\ldots \text{ }\left( i \right)\]

So,

    \[1.2\text{ }+\text{ }2.3\text{ }+\text{ }3.4\text{ }+\text{ }\ldots \text{ }+\text{ }k\left( k+1 \right)\text{ }+\text{ }\left( k+1 \right)\text{ }\left( k+2 \right)\]

Now, substituting the value of P (k) we get,

    \[=\text{ }\left[ k\text{ }\left( k+1 \right)\text{ }\left( k+2 \right) \right]\text{ }/\text{ }3\text{ }+\text{ }\left( k+1 \right)\text{ }\left( k+2 \right)\]

by using equation (i)

    \[=\text{ }\left( k+2 \right)\text{ }\left( k+1 \right)\text{ }\left[ k/2\text{ }+\text{ }1 \right]\]

    \[=\text{ }\left[ \left( k+1 \right)\text{ }\left( k+2 \right)\text{ }\left( k+3 \right) \right]\text{ }/3\]

P (n) is true for

    \[n\text{ }=\text{ }k\text{ }+\text{ }1\]

Hence, P (n) is true for all n ∈ N.

i

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