$1 + 2 + 3 + 4 + . . . . . + n = ?$
$1 + 2 + 3 + 4 + . . . . . + (n - 1) = ?$
$1 + 1 + 1 + 1 + . . . . . + n = ?$
$1 + 1 + 1 + 1 + . . . . . . + (n - 1) = ?$
$1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + n^{2} = ?$
$1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + {(n - 1)}^{2} = ?$

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Solution

$1 + 2 + 3 + 4 + . . . . . + n = \frac{n (n + 1)}{2}$
$1 + 2 + 3 + 4 + . . . . . + (n - 1) = \frac{n (n - 1)}{2}$
$1 + 1 + 1 + 1 + . . . . . + n = n$
$1 + 1 + 1 + 1 + . . . . . . + (n - 1) = (n - 1)$
$1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$
$1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + {(n - 1)}^{2} = \frac{(n - 1) (n) (2 n - 1)}{6}$

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Similar questions

Q. If

\frac{1 \times 2^{2} + 2 \times 3^{2} + 3 \times 4^{2} + \dots + n (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + 3^{2} \times 4 + \dots + n^{2} (n + 1)} = \frac{8}{7}

, then

n =

Q. Show that

\frac{1 \times 2^{2} + 2 \times 3^{2} + . . . + n \times {(n + 1)}^{2}}{1^{2} \times 2 + 2^{2} \times 3 + . . . + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}

Show that

$\frac{1 \times 2^{2} + 2 \times 3^{2} + \dots \dots + n \times (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + \dots \dots + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}$

\frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + \dots + n

terms

=

1^{2} + 1 + 2^{2} + 2 + 3^{2} + 3 + \dots . . + n^{2} + n =

Join BYJU'S Learning Program

1+2+3+4+.....+n=?1+2+3+4+.....+(n−1)=?1+1+1+1+.....+n=?1+1+1+1+......+(n−1)=?12+22+32+42+....+n2=?12+22+32+42+....+(n−1)2=?

1+2+3+4+.....+n=n(n+1)21+2+3+4+.....+(n−1)=n(n−1)21+1+1+1+.....+n=n1+1+1+1+......+(n−1)=(n−1)12+22+32+42+....+n2=n(n+1)(2n+1)612+22+32+42+....+(n−1)2=(n−1)(n)(2n−1)6

$1 + 2 + 3 + 4 + . . . . . + n = ?$
$1 + 2 + 3 + 4 + . . . . . + (n - 1) = ?$
$1 + 1 + 1 + 1 + . . . . . + n = ?$
$1 + 1 + 1 + 1 + . . . . . . + (n - 1) = ?$
$1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + n^{2} = ?$
$1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + {(n - 1)}^{2} = ?$