1) (x+2)^n-1 + (x+2)^n-2 (x+1) +(x+2)^n-3 (x+1)^2+..........(x+1)^n-1 equals to
a) (x+2)^n-1 - (x+1)^n b)(x+2)^n-1 -(x+1)^n c) none of these

2) Number of terms in a G.P. is even and sum of these terms is 5 times of sum of odd terms,then common ratio is
a) 2 b)3 c) 4 d) 5

1

Answers

2015-06-14T00:36:47+05:30

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(x+2)^{n-1}+ (x+2)^{n-2}* (x+1) +(x+2)^{n-3} *(x+1)^2 \\.\ \ \ \ \ ...+(x+1)^{n-1}\\\\let\ x+2=a,\ x+1=b\\\\So\ given\ \ a^{n-1}+ a^{n-2}* b +a^{n-3} *b^2+ ...+a\ b^{n-2}+b^{n-1}\\\\=(a+b)^{n-1}=(x+2+x+1)^{n-1}\\\\=(2x+3)^{n-1}
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number of terms = 2 n
T1 = a              common ratio = r
Geometric series:    a , a r, a r², .... a r^{2n-1}

sum\ S_{2n}=\frac{a(r^{2n+1}-1)}{r-1},\ \ ----(1)\\\\Series\ with\ odd\ terms:\ a, ar^2, ar^4...,ar^{2n-2}\\Common\ ratio=r^2\\\\Sum\ S_{n\ odd\ terms}=\frac{a((r^2)^{n+1}-1)}{r^2-1},\ \ ----(2)

Sum of even terms + sum of odd terms = sum of all terms = 5 * sum of odd terms
   =>  Sum of even terms = 4 * sum of odd terms 

even\ terms\ series:\ ar^2,ar^4,ar^6,...,ar^{2n-1}\\Common\ ratio=r^2,\ \ T_1=ar^2\\\\Sum\ S_{n\ even\ terms}=\frac{ar^2((r^2)^{n+1}-1)}{r^2-1}=\frac{ar^2(r^{2n+2}-1)}{r^2-1},\ \ \ ---(3)

\frac{S_{n\ even\ terms}}{S_{n\ odd\ terms}}=4=r^2

Hence,  r = +2 or -2

2 5 2
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