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Let a, b, c, be in A.P. ,with a common difference d, then e^{1/e} , e^{b}^{/ae}, e^{1/a} are in
Sum to 20 terms of the series (1.3)^{2} + (2.5)^{2} + (3.7)^{2} +… is:
Given that (1.3)^{2}, (2.5)^{2} , (3.7)^{2} .....................20 terms
Here 1,2,3,.................are in A.P.
a = 1, d = 1
t_{n} = a + (n1) d = 1 + (n 1)1 = n
3,5,7 ............. are in A.P
a = 3, d = 2
tn = a + (n1) d = 3 + (n 1)^{2} = 2n + 1
∴ nth term = n(2n + 1)^{2} = 4n^{3} + 4n^{2} + n
Sum of n terms Sn = ∑ ( 4n^{3} + 4n^{2} + n ) = 4 { n^{2}(n+1)^{2}}/4 + 4 {n (n+1)(2n+1)} / 6 + n(n+1)/2
= { n^{2}(n+1)^{2}} + 2 {n (n+1)(2n+1)} / 3 + n(n+1)/2
Sum of 20 terms S_{20} = 400 × 441 + 40 × 41 × 7 + 10 ×21 = 176400 + 11480 + 210
= 188090.
0.2 + 0.22 + 0.222 + 0.2222 + ….
2(0.1 + 0.11 + 0.111 + 0.1111 + ….)
2/9 (0.9 + 0.99 + 0.999 + 0.9999 + ….)
2/9 (( 1  0.1 ) + ( 1  0.01 ) + ( 1  0.001 ) + ( 1  0.0001 ) + …. )
2/9 (1  0.1 + 1  0.01 + 1  0.001 + 1  0.0001 + ….)
2/9 ((1 + 1 + 1 + 1 + ….)  ( 0.1 + 0.01 + 0.001 + 0.0001 + ….))
2/9 (n  ( 0.1 + 0.01 + 0.001 + 0.0001 + ….))
2/9 (n  (( 0.1 ( 1  0.1^{n})/( 1  0.1))))
2/9 (n  (( 0.1 / 0.9)( 1  0.1^{n})))
2/9 (n  (1/9)(1  0.1^{n}))
The sum of the series 3^{2} + 6^{2} + 9^{2} + … + (3n)^{2} is:
3^{2} + 6^{2 }+....
∑(3n)^{2}
9∑n^{2}
9n(n+1)(2n+1)/6
= 3n(n+1)(2n+1)/2
In how many different ways the letters of the word HEXAGON can be permuted?
The given word consists of 7 different letters out of which 3 are vowels and 4 are consonants.
7 different letters can be arranged in
7! ways = 5040 ways.
The sum of the squares of the first 15 natural numbers is
S_{n} = 1/6n(n+1)(2n+1)
Here n = 15 so,
S_{15} = 1/6 × 15(15+1)(2×15+1)
S_{15} = 1/6 × 15 × 16(30 + 1)
S_{15} = 1/6 × 15 × 16 × 31
S_{15} = 1240
Sum of squares of the first n natural numbers is given by:
The formula to find sum of first n terms of a sequence of squares of natural numbers = n(n+1)(2n+1)/6
The sum of the series 1^{3} – 2^{3} + 3^{3} – 4^{3} + ……. + 9^{3} is:
1^{3} + 3^{3} + 5^{3} + 7^{3} + 9^{3} − (2^{3} + 4^{3} + 6^{3} + 8^{3})
= Sum of cubes of first 5 odd number Sum of cubes of first 4 even number
= n^{2}(2n^{2}–1) − 2n^{2}(n+1)^{2}
= 1225 – 800
= 425
Sum to n terms of the series 2 + 5 + 14 + 41 +… is:
S_{n} = 2 + [2+3^{1}] + [2+3^{1}+3^{2}] + ⋯ + [2+3^{1}+3^{2}+⋯+3^{n1}]
= 2n+[3^{1}]+[3^{1}+3^{2}]+⋯+[3^{1}+3^{2}+⋯+3^{n1}]
= 2n + t_{3,1} + t_{3,2} +⋯+t_{3,n1}
= 2n+T_{3,n1}
= 2n+3/4(3^{n}−2n−1)
= n/2 + 3/4(3^{n} − 1)
If 1, a and P are in A. P. and 1, g and P are in G. P., then
2a = 1 + p and g^{2} = P
⇒ g^{2} = 2a − 1
⇒ 1 − 2a + g^{2} = 0
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