Binomial expansion: find T_3 and T_7 for (2x + 1/2)^7

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M4STR0k

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The question:

\(\displaystyle \mbox{In the expansion }\, \left(2x\, +\, \dfrac{1}{2}\right)^7,\, \mbox{ find }\, T_3\, \mbox{ and }\, T_7\, \mbox{ according to the decending power}\)

\(\displaystyle x,\, \mbox{ if }\, T_3\, =\, T_7\, \mbox{ Find }\, x.\)


The answer i got:

\(\displaystyle T_3\, =\, {}^7C_2\, (2x)^5\, \left(\dfrac{1}{2}\right)^2\, =\, {}^7C_2\, 32x^5\, \dfrac{1}{4}\)

\(\displaystyle T_7\, =\, {}^2C_6\, (2x)\, \left(\dfrac{1}{2}\right)^6\, =\, {}^7C_6\, 2x\, \dfrac{1}{64}\)

Any help ..
 

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M4STR0k said:
The question:

\(\displaystyle \mbox{In the expansion }\, \left(2x\, +\, \dfrac{1}{2}\right)^7,\, \mbox{ find }\, T_3\, \mbox{ and }\, T_7\, \mbox{ according to the decending power}\)

\(\displaystyle x,\, \mbox{ if }\, T_3\, =\, T_7\, \mbox{ Find }\, x.\)


The answer i got:

\(\displaystyle T_3\, =\, {}^7C_2\, (2x)^5\, \left(\dfrac{1}{2}\right)^2\, =\, {}^7C_2\, 32x^5\, \dfrac{1}{4}\)

\(\displaystyle T_7\, =\, {}^2C_6\, (2x)\, \left(\dfrac{1}{2}\right)^6\, =\, {}^7C_6\, 2x\, \dfrac{1}{64}\)

Any help ..
Click to expand...
What help do you need?
 
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M4STR0k said:
Is it okay or not ? want to check ...
Click to expand...
It's an acceptable start. Now finish the first part, and attempt the second part.

When you reply, please provide the full text of the exercise. Some of the text appears to have been cut off in your first image.

Thank you! ;)
 
M4STR0k said:
The question:

\(\displaystyle \mbox{In the expansion }\, \left(2x\, +\, \dfrac{1}{2}\right)^7,\, \mbox{ find }\, T_3\, \mbox{ and }\, T_7\, \mbox{ according to the decending power}\)

\(\displaystyle x,\, \mbox{ if }\, T_3\, =\, T_7\, \mbox{ Find }\, x.\)


The answer i got:

\(\displaystyle T_3\, =\, {}^7C_2\, (2x)^5\, \left(\dfrac{1}{2}\right)^2\, =\, {}^7C_2\, 32x^5\, \dfrac{1}{4}\)

\(\displaystyle T_7\, =\, {}^2C_6\, (2x)\, \left(\dfrac{1}{2}\right)^6\, =\, {}^7C_6\, 2x\, \dfrac{1}{64}\)

Any help ..
Click to expand...


So the next step would be:

\(\displaystyle {}^7C_2\, 32x^5\, \dfrac{1}{4} \ = \, {}^7C_6\, 2x\, \dfrac{1}{64}\)

\(\displaystyle {}^7C_2\, 16x^4\, \ = \, {}^7C_6\, \, \dfrac{1}{16}\)

\(\displaystyle {}^7C_2\, (4x)^4\, \ = \, {}^7C_6\,\)

and continue....
 
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