find-the-sum-of-the-first-n-terms-from-1-2-3-4-5-6-7-8-9-10-

Question Number 198243 by mr W last updated on 15/Oct/23

f i n d t h e s u m o f t h e f i r s t n t e r m s f r o m

1, 2 + 3, 4 + 5 + 6, 7 + 8 + 9 + 10, \dots

Answered by som(math1967) last updated on 15/Oct/23

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + . . n

n o o f t e r m = \frac{n (n + 1)}{2}

s u m o f \frac{n (n + 1)}{2} t e r m s

= \frac{\frac{n (n + 1)}{2} \times {\frac{n (n + 1)}{2} + 1}}{2}

= \frac{n (n + 1) (n^{2} + n + 2)}{8}

Answered by universe last updated on 15/Oct/23

1, 2 + 3, 4 + 5 + 6, 7 + 8 + 9 + 10, \dots = {a_{n}}

\frac{n^{3} + n}{2} \to t h i s i s a s u m o f n^{t h} t e r m o f

s e q u e n c e {a_{n}}

n o w

P (l e t) = \underset{n = 1}{\sum^{n}} (\frac{n^{3}}{2} + \frac{n}{2})

P = \frac{1}{2} \times {[\frac{n (n + 1)}{2}]}^{2} + \frac{1}{2} \times \frac{n (n + 1)}{2}

P = \frac{n (n + 1) (n^{2} + n + 2)}{8}

Answered by mr W last updated on 15/Oct/23

t h e n^{t h} t e r m h a s k_{n} = n n u m b e r s

\underset{i = 1}{\sum^{n - 1}} k_{i} = \frac{n (n - 1)}{2}

a_{n} = [\frac{n (n - 1)}{2} + 1] + [\frac{n (n - 1)}{2} + 2] + \dots + [\frac{n (n - 1)}{2} + n]

a_{n} = n \times \frac{n (n - 1)}{2} + 1 + 2 + \dots + n

a_{n} = n \times \frac{n (n - 1)}{2} + \frac{n (n + 1)}{2} = \frac{n (n^{2} + 1)}{2}

s_{n} = \underset{k = 1}{\sum^{n}} a_{k} = \underset{k = 1}{\sum^{n}} \frac{k (k^{2} + 1)}{2} = \frac{1}{2} \underset{k = 1}{\sum^{n}} (k + k^{3})

= \frac{1}{2} [\frac{n (n + 1)}{2} + \frac{n^{2} {(n + 1)}^{2}}{4}]

= \frac{n (n + 1) (n^{2} + n + 2)}{8} ✓

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