Find-the-sum-of-the-cubes-of-first-n-even-number-and-first-n-odd-number-

Question Number 57635 by Tawa1 last updated on 09/Apr/19

Find the sum of the cubes of first n even number, and

first n odd number .

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19

A s p e r h i g h e r a l g e b r a B e r n a r d a n d c h i l d

w e c a n f i n d

s_{r} = 1^{r} + 2^{r} + 3^{r} + \dots + n^{r}

b y i n t r e g a t i o n s n d i n t r e g a t i o n c o n s t a n t

a s B e r n u l i n u m b e r s

e x a m p l e

s_{1} = 1 + 2 + 3 + \dots + n

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

n o w t o f i n d S_{2} = 1^{2} + 2^{2} + 3^{2} + \dots + n^{2}

u s i n g S_{1}

S_{r} = r \int S_{r - 1} d n + {n B}_{r} \leftarrow f o r m u l a

S_{2} = 2 \int (\frac{n^{2}}{2} + \frac{n}{2}) + n \times \frac{1}{6} [B_{2} = \frac{1}{6}]

= 2 (\frac{n^{3}}{6} + \frac{n^{2}}{4}) + n \times \frac{1}{6}

= \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6}

= \frac{2 n^{3} + 3 n^{2} + n}{6}

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

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

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

s i m i l a r l y

S_{3} = 3 \int S_{2} d n + {n B}_{3}

= 3 \int (\frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6}) d n + n \times 0

= \frac{n^{4}}{4} + \frac{n^{3}}{2} + \frac{n^{2}}{4}

= \frac{n^{4} + 2 n^{3} + n^{2}}{4}

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

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

t h u s y o u c a n f i n d \dots

Commented by Tawa1 last updated on 09/Apr/19

God bless you sir . I appreciate .

Answered by mr W last updated on 09/Apr/19

S_{E} = \underset{k = 1}{\sum^{n}} {(2 k)}^{3} = 8 \underset{k = 1}{\sum^{n}} k^{3} = 8 \times \frac{n^{2} {(n + 1)}^{2}}{4}

= 2 n^{2} {(n + 1)}^{2}

S_{O} = \underset{k = 1}{\sum^{n}} {(2 k - 1)}^{3} = \underset{k = 1}{\sum^{n}} (8 k^{3} - 12 k^{2} + 6 k - 1) = 8 \times \frac{n^{2} {(n + 1)}^{2}}{4} - 12 \times \frac{n (n + 1) (2 n + 1)}{6} + 6 \times \frac{n (n + 1)}{2} - n

= n^{2} (2 n^{2} - 1)

Commented by Tawa1 last updated on 09/Apr/19

Wow, God bless you sir . I appreciate .

Commented by Tawa1 last updated on 09/Apr/19

But sir, i have a question .

Please see this question .

simplify : \frac{\sqrt{10 + \sqrt{1}} + \sqrt{10 + \sqrt{2}} + \dots + \sqrt{10 + \sqrt{99}}}{\sqrt{10 - \sqrt{1}} + \sqrt{10 - \sqrt{2}} + \dots + \sqrt{10 - \sqrt{99}}}

Answer : \sqrt{2} + 1

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19

e x c e l l e n t s i r \dots

Commented by Tawa1 last updated on 09/Apr/19

God bless you sir . But i have checked . i usually undersand your

solution better sir .

Commented by mr W last updated on 09/Apr/19

i l i k e t h i s s o l u t i o n :

A = \underset{k = 1}{\sum^{99}} \sqrt{10 + \sqrt{k}}

B = \underset{k = 1}{\sum^{99}} \sqrt{10 - \sqrt{k}}

{(\sqrt{10 + \sqrt{k}} - \sqrt{10 - \sqrt{k}})}^{2} = 10 + \sqrt{k} - 2 \sqrt{(10 + \sqrt{k}) (10 - \sqrt{k})} + 10 - \sqrt{k}

{(\sqrt{10 + \sqrt{k}} - \sqrt{10 - \sqrt{k}})}^{2} = 2 (10 - \sqrt{100 - k})

\Rightarrow \sqrt{10 + \sqrt{k}} - \sqrt{10 - \sqrt{k}} = \sqrt{2} \sqrt{10 - \sqrt{100 - k}}

\Rightarrow \underset{k = 1}{\sum^{99}} \sqrt{10 + \sqrt{k}} - \underset{k = 1}{\sum^{99}} \sqrt{10 - \sqrt{k}} = \sqrt{2} \underset{k = 99}{\sum^{1}} \sqrt{10 - \sqrt{100 - k}}

\Rightarrow A - B = \sqrt{2} B

\Rightarrow A = (\sqrt{2} + 1) B

\Rightarrow \frac{A}{B} = \sqrt{2} + 1

Commented by Tawa1 last updated on 09/Apr/19

God bless you sir . I appreciate .

Commented by Tawa1 last updated on 09/Apr/19

Sir, you said B = \underset{k = 1}{\sum^{99}} \sqrt{10 - \sqrt{k}}

finally how is \underset{k = 99}{\sum^{1}} \sqrt{10 - \sqrt{100 - k}} = B again \dots

Commented by mr W last updated on 09/Apr/19

g e n e r a l l y :

\frac{\underset{k = m}{\sum^{n^{2} - m}} \sqrt{n + \sqrt{k}}}{\underset{k = m}{\sum^{n^{2} - m}} \sqrt{n - \sqrt{k}}} = \sqrt{2} + 1 w i t h n, m \in N

Commented by Tawa1 last updated on 09/Apr/19

wow . I understand now . God bless you .

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19

l o o k \dots B (w h e n k \to 1 \to 99) = \underset{k = 1}{\sum^{99}} \sqrt{10 - \sqrt{k}}

B (k \to 1 \to 99 = \sqrt{10 - \sqrt{1}} + \sqrt{10 - \sqrt{2}} + \sqrt{10 - \sqrt{3}} + . . + \sqrt{10 - \sqrt{99}}

n o w l o o k

B (k \to 99 \to 1) [\underset{k = 99}{\sum^{1}} \sqrt{10 - \sqrt{100 - k}}]

= \sqrt{10 - \sqrt{1}} + \sqrt{10 - \sqrt{2}} + \dots + \sqrt{10 - \sqrt{99}}

h e n c e

\underset{k = 1}{\sum^{99}} \sqrt{10 - \sqrt{k}} = \underset{k = 99}{\sum^{1}} \sqrt{10 - \sqrt{100 - k}}

Commented by Meritguide1234 last updated on 10/Apr/19

see q . 54226 (ans already provided)

Commented by malwaan last updated on 10/Apr/19

thank you

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