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Artemon [7]
3 years ago
8

Use​ Gauss's approach to find the following sums​ (do not use​ formulas). a. 1+2+3+4+...+98

Mathematics
1 answer:
Blababa [14]3 years ago
6 0
Here ya go im 90% sure this is the right answer enjoy!

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3 years ago
Expand (2x+2)^6<br> How would you find the answer using the binomial theorem?
Yanka [14]

Answer:

Step-by-step explanation:

\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.

\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\

(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+  \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.

8 0
1 year ago
I need help pls a s a p
jonny [76]

Answer:

could it be as simple as x+1=y or y-1=x?

Step-by-step explanation:

y is all of x plus one more.....

7 0
2 years ago
Nguyen played a game in which he could either lose or gain points each round. at the end of 7 rounds, he had -8 points. After on
Sergeeva-Olga [200]

Answer:

I think the answer would be +15.

Step-by-step explanation:

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2 years ago
A family paid $142.80 which included $6.80 in taxes for 4 play tickets each ticket costs the same amount how much did each perso
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$142.80-$6.80=$136 divided by 4 equals $34
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3 years ago
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