Solution:
Σ5r=1 r2 = 5(5+1)(2.5+1)/6 = 55
Illustration:
Evaluate: 62 + 72 + 82 + 92 + 102
Solution:
Required sum = 12 + 22 + 32 + …… +102 – (12 + 22 +…… 52)
= Σ10r=1 r2 – Σ5r=1 r2
= 10(10+1)(2.10+1)/6 – 5(5+1)(2.5+1)6
= 385 – 55 = 330
Illustration:
Evaluate : 12 + 32 + 52 +………+ (2n–1)2
Solution:
Note:
There are n in terms in the series.
Required sum = 12 + 22 + 32 + 42 +……+ (2n–1)2 + (2n)2
–(22 + 42 +………+ (2n)2)
= (2n)(2n+1)(4n+1)/6 – 4 (12 + 22 + ……+ n2)
= (2n)(2n+1)(4n+1)/6 – 4 . n(n+1)(2n+1)/6
=n(4n2–1)/3
Illustration:
Sum: 1 . 2 + 2 . 3 + 3 . 4 +…… + n.(n+1)
Solution:
tr = r(r+1)
= r2 + r
t1 = 12 + 1
t2 = 22 + 2
t3 = 32 + 3
tn = n2 + n
Adding
t1 = t2 + t3 +…+ tn = (12 + 22 + 32 +…+ n2)+(1+2+3+…+n).
Σnr=1 tr = Σnr=1 r2 + Σnr=1 r
Sn = n(n+1)(2n+1)/6 + n(n+1)/2
= n(n+1)/6 [2n + 1 + 3]
= n(n+1)/6 (2n + 4)
=n(n+1)(n+2)/3
Illustration:
Sum to n-terms =1/1.2 + 1/2.3 + 1/3.4 +......+ 1/n(n+1)
Solution:
tr = 1/r(r+1)
= 1/r – 1/r+1
t1 = 1 –1/2
t2 = 1/2 – 1/3
t3 = 1/3 – 1/4
t4 = 1/4 – 1/5
tn =1/n – 1/n+1
Adding
Σnr=1 tr = 1 – 1/n+1 = n/n+1
Factorial:
Factorial of a natural number n is defined as the product of first n natural numbers and it is noted ên or n!
|0 = 1 (by definition)
|1 = 1
|2 = 2. |1 = 2
|3 = 3 . |2 = 3.2.1 = 6
|4 = 4 . |3 = 4.3.2.1 =24
|n = n |(n-1)
Illustration:
Evaluate the sum : 1 |1 + 2 |2 + 3 |3 +… to n terms.
Solution:
tr = r |r
= |r + 1 – |r
t1 = |2 – |1
t2 = |3 – |2
t3 = |4 – |3
tn = |n + 1 – |n
————————————
Σnr=1 tr = |n + 1 – 1
Illustration:
If x, y, z are positive real numbers, such that x + y + z = a, then prove that 1/x + 1/y + 1/z > 9/a.
Solution:
Since A.M. > H.M.
x+y+z/3 > 3/x–1+y–1+z–1 ⇒ 1/x + 1/y + 1/z > 9/a.
Illustration:
Prove that (a + b + c) (ab + bc + ca) > 9abc.
Solution:
Using AM > GM, we have
a+b+c/3 > (abc)1/3 and ab+bc+ca/3 > (ab.bc.ca)1/3.
Multiplying these two results, we have
(a+b+c/3)(ab+bc+ca/3) >(abc)1/3 (ab.bc.ca)1/3
or,(ab+bc+ca)(a+b+c)/9 > (a3b3c3)1/2
or, (a + b + c)(ab + bc + ca) > 9abc.
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