Prove by mathematical induction that
n. 1 + ( n - 1) 2 + (n - 2) 3 + ..... + 2 (n -1) + 1.n = $\frac{n}{6}$ (n + 1)(n + 2)

True

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False

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Solution

The correct option is A True
Result is obviously true for $n = 1$ Assume P(n) i.e.
$P (n) = n \cdot 1 + (n - 1) \cdot 2 + . . . . + 2 (n - 1) + + 1 \cdot n$
$= \frac{n}{6} (n + 1) (n + 2)$
p(n + 1) = (n + 1) 1 + n 2+ ,... 3(n -1)+2n + 1(n +1)
$∴ p (n + 1) - p (n) = 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 + 1 (n - 1) + 1 \cdot n + 1 + 1 \cdot (n + 1)$
= 1 + 2 + 3 ... + n + 1(n + 1)
$= \frac{n (n + 1)}{2} + n + 1 = \frac{(n + 1) (n + 2)}{2}$
$∴ p (n + 1) = \frac{(n + 1) (n + 2)}{2} p (n)$
$\frac{(n + 1) (n + 2)}{2} + \frac{(n + 1) (n + 2)}{6}$
$\frac{(n + 1) (n + 2)}{2} 1 + [\frac{n}{3}]$
$\frac{(n + 1) (n + 2) (n + 3)}{6} = \frac{n}{6} (n + 1) (n + 2)$
Thus p (n + 1) also holds good

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Prove by mathematical induction that n. 1 + ( n - 1) 2 + (n - 2) 3 + ..... + 2 (n -1) + 1.n = n6 (n + 1)(n + 2)

Prove by mathematical induction that
n. 1 + ( n - 1) 2 + (n - 2) 3 + ..... + 2 (n -1) + 1.n = $\frac{n}{6}$ (n + 1)(n + 2)