From principle of mathematical induction prove that
$1 + 2 + 3 + . . . . + n = \frac{1}{2} n (n + 1)$

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Solution

$1 + 2 + 3 n$
$A P$
Sum of $n$ term $= \frac{n}{2} [2 a + (n - 1) d]$
$= (\frac{n}{2}) [2 (a) + (n - 1) 1]$
$= \frac{n}{2} [n + 1]$

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From principle of mathematical induction prove that1+2+3+....+n=12n(n+1)

1+2+3nAPSum of n term =n2[2a+(n−1)d]=(n2)[2(a)+(n−1)1]=n2[n+1]

From principle of mathematical induction prove that
$1 + 2 + 3 + . . . . + n = \frac{1}{2} n (n + 1)$

$1 + 2 + 3 n$
$A P$
Sum of $n$ term $= \frac{n}{2} [2 a + (n - 1) d]$
$= (\frac{n}{2}) [2 (a) + (n - 1) 1]$
$= \frac{n}{2} [n + 1]$