B1A4的《What's Going On》的歌詞如下：
What's going on
Yo let me holla at you
Let me see y'all niggaz say it
Yeah yeah yeah oh oh oh
What's going on wit my homeboys
Let me hear the ruckus roar
Everybody move your body
Swing that ass up here and throw your hands up in the air
Get a whoop that round
If your girl jump off lets have it go off oh yeah oh yeah oh oh oh
See they go head let the mothafuckin good times roll on roll on baby oh yeah oh yeah oh oh oh
Let me see y'all niggaz say it Yeah yeah yeah oh oh oh
What's going on with my homeboys let me hear the ruckus roar
Everybody say it with me one time Now who feel it let me hear you sound Let me see you bouncin back back back
Come back like I know your out in the streets tryna catch some bread now Everybody get low low low Hey hey hey yo Yo and let it flow like so good yo and just pop the top pop top and pass that no tops we gotta grab the bag we get so mad Hey yo it's my time of the month you don't stop I'm in the zone I'm ballin hard I'm ballin hard Hey yo I ain't even gotta call my homies up they gon come they gon come Hey hey hey we gon hit the block and get some smoke and get loose Hey hey hey we ain't stoppin till we see the bottom Hey hey hey and then we bounce bounce bounce Hey hey hey let me see y'all niggaz say it Yeah yeah yeah oh oh oh Hey hey hey we ride for them big wheels Hey hey hey and them big speakers Hey hey hey when them big guns come through Hey hey hey I ain't tryna play no games like last time Now where them fuckin whip jaws at where at at at Yep that about it cuz a young dog like me in it let me hear you scream like設正項等比數列｛a\n｝的前n項和為Sn，若a7是a4與a12的等比中項，求數列｛a\n｝的公比q及數列｛Sn｝的前n項和。
求過程，謝謝！
【分析】
由等比數列的性質得$a_{4} \cdot a_{12} = a_{7}^{2}$，再由等比數列的求和公式即可求出公比$q$及數列｛$S_{n}$｝的前$n$項和．
【解答】
解：由題意得$a_{7}^{2} = a_{4} \cdot a_{12}$，即$a_{4} \cdot a_{12} = {a_{7}}^{2}$，又因為｛$a_{n}$｝是正項等比數列，所以$q^{3} = 2$，所以$q = \sqrt[3]{2}$．
所以$S_{n} = \frac{a_{1}(1 - q^{n})}{1 - q}$，當$n = 1$時，$S_{1} = \frac{a_{1}}{1 - q}$，當$n \geqslant 2$時，$S_{n} = \frac{a_{1}(1 - q^{n})}{1 - q} - \frac{a_{1}(1 - q^{n - 1})}{1 - q}$，所以數列｛$S_{n}$｝的前$n$項和為$\frac{a_{1}}{1 - q} + \frac{a_{1}(q^{n} - 1)}{(q - 1)^{2}}$．