[SOLVED] Binomial Theorem help...

**looi76** · February 1st 2008, 07:56 AM

Question:
Expand $(3+5x)^7$ in ascending powers of $x$ up to and including the term in $x^2$ . By putting $x=0.01$ , find an approximation, correct to the nearest whole number, to $3.05^7$ .

**topsquark** · February 1st 2008, 09:09 AM

Originally Posted by looi76

Question:
Expand $(3+5x)^7$ in ascending powers of $x$ up to and including the term in $x^2$ . By putting $x=0.01$ , find an approximation, correct to the nearest whole number, to $3.05^7$ .

You've asked for help on a few of these now. What part of it is confusing you? Tell us so we can help you better.
$(3 + 5x)^7 = \sum_{k = 0}^7 {7 \choose k} (3)^{7 - k}(5x)^k = 3^7 + 7 \cdot (3)^6 (5x)^1 + ~...$

You tell me what the next term is.

-Dan

**colby2152** · February 1st 2008, 09:17 AM

Ultimately, in that approximation of $3.05^7$ , you will have to calculate some $(.01a)^b$ for seven different sets of $a$ and $b$ .

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