Find the term independent of x in the expansion of ( 1 + x + x^-3)^7. Hint: Treat 1 + x as a single term.
Somehow, I can't find a way to group the 'x' together because of the addition sign. I only need someone to tell me how to group the x into one so as to equal it to x^0. Thanks.
what + sign
Post what you have done here, then we will try to correct it. It's a better way to learn
(7Cr)(1 + x)^7-r (x^-3)^r
Usually we should be able to group the x together. But I can't manage to do so here. Yea, so I'm stuck here.
expand the (1+x)^7 to the general term again
Huh, why is it (1+x)^7 ? Cuz it is already (1+x)^7-r when you expand to the general term in the first place.
Hi.
Been trying this question bonky here has posted. Sorry to say, I still can't solve it. ROFL.......... eh, bonky, will try to ask around within my circle of friends. Get back to you ASAP yea.
Sorry. Still can't manage to find the solution leh. Perhaps you want to try asking others? Guys, do give a hand if you know how to solve this. (:
oops.. now then I see it
give me a few minutes... a bit busy with increasing number of assignments lately
Find the term independent of x in the expansion of ( 1 + x + x^-3)^7. Hint: Treat 1 + x as a single term.
( 1 + x +
x^-3)^7
= ( (1 + x) +
x^-3)^7
General Term = 7Cr (1+x)^(7-r) (x^-3)^r, where r is from 0 to 7
General Term of (1+x)^(7-r) = (7-r)Cq (x)^q
So, general term = 7Cr (7-r)Cq (x)^q (x^-3)^r
When r = 0, q = 0, thus independent term = 1
When r = 1, q = 3, thus independent term = 7C1 * 6C3 = 140
When r = 2, q = 6, thus independent term = 7C2 * 6C6 = 21
When r = 3 to 7, there's no independent term
Hence, term independent of x in the expansion of ( 1 + x + x^-3)^7 is 1 + 140 + 21 = 162