Question-164206

Question Number 164206 by abdurehime last updated on 15/Jan/22

Commented by abdurehime last updated on 15/Jan/22

what is the value ? ? ?

Commented by mr W last updated on 15/Jan/22

1^{2} + 2^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}

Commented by abdurehime last updated on 15/Jan/22

how did you get ? ? show me the process please ?

Commented by cortano1 last updated on 15/Jan/22

c o n s i d e r {(k + 1)}^{2} - k^{2} = 2 k + 1

\underset{k = 1}{\sum^{n}} ({(k + 1)}^{2} - k^{2}) = \underset{k = 1}{\sum^{n}} (2 k + 1)

{(n + 1)}^{2} - 1 = 2 \underset{k = 1}{\sum^{n}} k + n

n^{2} + n = 2 \underset{k = 1}{\sum^{n}} k \Rightarrow \underset{k = 1}{\sum^{n}} k = \frac{n^{2} + n}{2}

{(k + 1)}^{3} - k^{3} = 3 k^{2} + 3 k + 1

\underset{k = 1}{\sum^{n}} ({(k + 1)}^{3} - k^{3}) = 3 \underset{k = 1}{\sum^{n}} + 3 \underset{k = 1}{\sum^{n}} + n

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

3 \underset{k = 1}{\sum^{n}} k^{2} = n^{3} + 3 n^{2} + 3 n - (\frac{3 n^{2} + 5 n}{2})

\underset{k = 1}{\sum^{n}} k^{2} = \frac{2 n^{3} + 3 n^{2} + n}{6} = \frac{n (2 n + 1) (n + 1)}{6}

Commented by bobhans last updated on 16/Jan/22

nice!

Answered by MJS_new last updated on 15/Jan/22

one possible process :

s_{n} = \underset{j = 1}{\sum^{n}} j^{2}

make a list of the first few s_{n}

write σ_{n}^{(1)} = s_{n + 1} - s_{n} below

write σ_{n}^{(2)} = σ_{n + 1}^{(1)} - σ_{n}^{(1)} below

continue until you get a constant (σ_{n}^{(k)} = σ_{n + 1}^{(k)})

s 1 5 14 30 55

σ^{(1)} 4 9 16 25

σ^{(2)} 5 7 9

σ^{(3)} 2 2

k = 3 \Rightarrow {we}^{'} ll get a polynome of 3^{rd} degree

s_{n} = c_{3} n^{3} + c_{2} n^{2} + c_{1} n + c_{0}

k + 1 = 4 constants \Rightarrow we need 4 equations

1 = c_{3} + c_{2} + c_{1} + c_{0}

5 = 8 c_{3} + 4 c_{2} + 2 c_{1} + c_{0}

14 = 27 c_{3} + 9 c_{2} + 3 c_{1} + c_{0}

30 = 64 c_{3} + 16 c_{2} + 4 c_{1} + c_{0}

solve this system \Rightarrow

c_{3} = \frac{1}{3} \land c_{2} = \frac{1}{2} \land c_{1} = \frac{1}{6} \land c_{0} = 0

\Rightarrow

s_{n} = \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} = \frac{n (n + 1) (2 n + 1)}{6}

Commented by Rasheed.Sindhi last updated on 16/Jan/22

G_{S \underset{!}{I} R}^{R^{E} A} T

Answered by ajfour last updated on 16/Jan/22

\underset{r = 1}{\sum^{n}} r (r + 1) (r + 2) . . (r + m)

= \frac{n (n + 1) (n + 2) \dots (n + m) (n + m + 1)}{m + 2}

h e r e Σ r^{2} = Σ r (r + 1) - Σ r

= \frac{n (n + 1) (n + 2)}{3} - \frac{n (n + 1)}{2}

= \frac{n (n + 1)}{6} {2 (n + 2) - 3}

\Rightarrow \underset{r = 1}{\sum^{n}} r^{2} = \frac{n (n + 1) (2 n + 1)}{6}

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