IIT JEE , AIEEE Forum - Mathematics , Algebra

shailesh_45

1] If 1+ 3+ 9 + 27+...to n terms exceeds 1000, find the minimum value of n.

2] Find the sum to 2n terms of the series whose every even term is a

times the term before it and every odd term is c times the term

beforeit, the first term being unity.

3] Find the sum of the proper divisor of the fillowing

(i) 2^10 .5^7. 11⁸ii) 6^a . 10^b . 15^c

where a,b , c are positive integers

risin

1)In gp,so 1(1-3ⁿ)/1-3 >=1000

3ⁿ-1>=2000

3ⁿ>=2001

=>n=7.

mujtaba4iit

a)S_n=a(rⁿ-1)/(r-1)

Sn= 1(3ⁿ-1)/(3-1)
Sn= (3ⁿ-1)/2

Now since u want the value of n so that the term just exceeds 1000.
1000<(3ⁿ-1)/2
2000<(3ⁿ-1)
2001<3ⁿ
Now we shall see the value of n for which the value of 3ⁿ is more than 1000.

3⁶ =729
3⁷ =2187 which is the minimum for which the value is more than 1000.
so, the answer is 7

Plz rate if its helpful and correct.

mukundmadhav

3) Consider the expression (1+ 2 + 2^2 + ..... 2^10)(1+ 5+ 5^2 +....5^7)(1+ 11+ 11^2+ .... 11^8)

Each term of the expansion of this will be a divisor of the given number, and no two terms will be the same as 2, 5, 11 are prime numbers.. So the given expression is the sum you're looking for. Solve the inner brackets by expression for sum of a GP
S=a(r^n-1)/(r-1) and multiply..

Rate if helpful..

spideyunlimited

2] Find the sum to 2n terms of the series whose every even term is a times the term before it and every odd term is c times the term before it, the first term being unity.

T1 T2 T3.......Tn......T2n

T1 = 1
T2 = a
T3 = ca
T4 = c.a^2
T5 = c^2.a^2
Tk if k is even = a^(k/2). c^(k/2 - 1)

T2n = a^(2n/2).c^(2n/2 -1)
T2n = a^n. c^(n-1)

S 2n = 1 + a + ca + c.a^2 + c^2.a^2 + c^2.a^3 .....a^n. c^(n-1)
= 1 + [ a + ca^2 + c^2.a^3 ....+ a^n.c^(n-1) ] + [ ca + c^2.a^2 + c^3.a^3..... + a^(n-1). c^(n-1) ]
= 1 + [ a.(a^n.c^n - 1) / (ac - 1) ] + [ ac( a^(n-1).c^(n-1) - 1) / (ac - 1) ]
..
..
.. solving further..

---> S 2n = (a^n.c^n - 1)(a + 1) / (ac - 1)

spideyunlimited

correct?

allamraju

2)The sum is 1+a+ac+a²c+a²c²+a³c²+a³c³+......upto 2n terms=(1+a)+ac(1+a)+a²c²(1+a)+....upto n terms.This is a G.P with first term (1+a) and common ratio ac.Hence,the sum is

(1+a)[1-(ac)ⁿ]/(1-ac)

allamraju

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2)The sum is 1+a+ac+a²c+a²c²+a³c²+a³c³+......upto 2n terms=(1+a)+ac(1+a)+a²c²(1+a)+....upto n terms.This is a G.P with first term (1+a) and common ratio ac.Hence,the sum is

(1+a)[1-(ac)ⁿ]/(1-ac)

3)If p₁,p₂,... are primes then the sum of proper divisors of p₁^k1.p₂^k2......is

[(p₁^k1+1-1)/p₁-1][p₂^k2+1-1/p₂-1]......

So,for the first bit,It is (2¹¹-1)(5⁸-1/4)(11⁹-1/10)

for the second bit,Write it as 2^a+b.3^a+c.5^b+c and apply the above formula

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2)The sum is 1+a+ac+a2c+a2c2+a3c2+a3c3+......upto 2n terms=(1+a)+ac(1+a)+a2c2(1+a)+....upto n terms.This is a G.P with first term (1+a) and common ratio ac.Hence,the sum is

(1+a)[1-(ac)n]/(1-ac)

3)If p1,p2,... are primes then the sum of proper divisors of p1k1.p2k2......is

[(p1k1+1-1)/p1-1][p2k2+1-1/p2-1]......

So,for the first bit,It is (211-1)(58-1/4)(119-1/10)

for the second bit,Write it as 2a+b.3a+c.5b+c and apply the above formula

2)The sum is 1+a+ac+a²c+a²c²+a³c²+a³c³+......upto 2n terms=(1+a)+ac(1+a)+a²c²(1+a)+....upto n terms.This is a G.P with first term (1+a) and common ratio ac.Hence,the sum is

(1+a)[1-(ac)ⁿ]/(1-ac)

3)If p₁,p₂,... are primes then the sum of proper divisors of p₁^k1.p₂^k2......is

[(p₁^k1+1-1)/p₁-1][p₂^k2+1-1/p₂-1]......

So,for the first bit,It is (2¹¹-1)(5⁸-1/4)(11⁹-1/10)

for the second bit,Write it as 2^a+b.3^a+c.5^b+c and apply the above formula