Question 1184224
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The coefficients that are in arithmetic progression can be found algebraically; however, it is unclear what the question is asking for....<br>
The coefficients in the expansion of (1+x)^n are...<br>
1st term: {{{1}}}
2nd term: {{{n}}}
3rd term: {{{((n)(n-1))/2}}}
4th term: {{{((n)(n-1)(n-2))/6}}}<br>
The 2nd, 3rd, and 4th terms are in arithmetic progression, so the 3rd term is twice the sum of the 2nd and 4th terms:<br>
{{{2((n)(n-1))/2=n+((n)(n-1)(n-2))/6}}}
{{{n(n-1)=n(1+((n-1)(n-2))/6)}}}
{{{n-1=1+(n^2-3n+2)/6}}}
{{{n-2=(n^2-3n+2)/6}}}
{{{6n-12=n^2-3n+2)}}}
{{{n^2-9n-14=0}}}
{{{(n-7)(n-2)=0}}}<br>
n=7 or n=2<br>
n=2 makes no sense in the problem because (1+x)^2 only has 3 terms.<br>
So n=7.<br>
The 2nd, 3rd, and 4th terms in the expansion of (1+x)^7 are 7, 21, and 35 -- in arithmetic progression with common difference 14.<br>
But I don't know how to answer the question because I don't know what it means:<br>
"Find the number of coefficients of the odd powers of x in the expansion."<br>
The expansion of (1+x)^7 has 8 terms; half of them are even powers of x and half are odd powers.<br>
So I guess the answer is 4.<br>
But after doing some good math to find a value of n for which the coefficients of the 2nd, 3rd, and 4th terms in the expansion of (1+x)^n form an arithmetic progression, that seems like a very odd question to ask....<br>