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3 years ago

Find the sum of the first 5 terms of a GP 1/7 + 2/7 + 4/7

Mathematics

1 answer:

Stels [109]3 years ago

7 0

Answer:

2 is the value of the infinite sum.

Step-by-step explanation:

3/2 = 2 - 1/2 7/4 = 2 - 1/4 15/8 = 2 - 1/8 and so on; the sum is 2

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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

Find the number that makes the ratio equivalent to 3:9. 9:?

sesenic [268]

Answer:

Step-by-step explanation:

Given the ratio 3 : 9

and 9 : ?

note that 3 × 3 = 9, thus 9 × 3 = 27

3 : 9 = 9 : 27

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3 years ago

Area of square is numerically less than twice its diagonal by 2. Find its Perimeter.

Usimov [2.4K]

2D-a^2=2

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3 years ago

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krek1111 [17]

Answer:

Step-by-step explanation:

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3 years ago

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7nadin3 [17]

Answer: The answer is the other guys thing

Step-by-step explanation:

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3 years ago

Find the sum of the first 5 terms of a GP 1/7 + 2/7 + 4/7​

Find the sum of the first 5 terms of a GP 1/7 + 2/7 + 4/7