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Using binomial theorem show that 1^(99) ...

Using binomial theorem show that `1^(99) + 2^(99) +3^(99) + 4^(99) + 5^(99)` is divisible by 5

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`1^(99) + 2^(99) + 3^(99) + 4^(99) + 5^(99)`
= 1+ (5-3)^(99) + 3^(99) + (5-1)^(99) + 5^(99)
= 1 + (5^99- "^(99)C_1 5^(98). 3^1 + ^(99)C_2 5^(97). 3^2 - ... 3^(99)) + 3^99 - (1- ^(99)C_1 5^1 + ^(99)C_2 5^2 - ... - 5^(99)) + 5^(99)`
= `(3xx5^(99) - "^(99)C_1 5^(98). 3^1 + ^(99)C_2 5^(97). 3^2 - .... + ^(99)C_(98) 5^1. 3^(98)) + (^(99)C_15^1 - ^(99)C_2 5^2 + ... -^(99)C_98 5^(98)) ... (1)
which is divisible by 5 as each term is a multiple of 5
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MBD PUBLICATION-Elements of Mathematics-QUESTION BANK
  1. Prove that "^(2n)C1 + ^(2n)C3 + .... + ^(2n)C(2n-1) = 2^(2n-1)

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  2. Find the sum of C1 + 2C2 + 3C3 + .... + nCn

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  3. Find the sum of C0 + 2C1 + 3C2 + .... + (n+1)Cn

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  4. Compute ((1+k)(1+k/2) ..... (1+k/n))/((1+n)(1+n/2) ..... (1+n/k))

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  5. Show that C0 C1 + C1 C2 + C2 C3 + .... + C(n-1) Cn = (2n!)/((n-1)!(n...

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  6. C0 C1 + C1 C2 + .... + C(n-1) Cn = (2^n.n.1.3.5... (2n-1))/(n+1)

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  7. Show that 3C0-8C1 + 13C2 - 18C3 + ..... + (n+1)^(th) term = 0

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  8. Show that C0 n^2 + C1 (2-n)^2 + C2 (4-n)^2 + .... + Cn (2n-n)^2 = n.2^...

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  9. Show that C0 + 3C1 + 5C2 + .... +(2n+1) Cn = (n+1)(2^n)

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  10. Find the sum of the following C1 - 2C2 + 3C3 - ..... + n(-1)^(n-1) Cn

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  11. Find the sum of the following 1.2 C(2) + 2.3 C(3) + ... + (n-1)nCn

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  12. Find the sum of the following C1 + 2^2 C2 + 3^2 C3 + ... + n^2Cn

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  13. Find the sum of the following C1 - 2C2 + 3C3 - ..... + n(-1)^(n-1) Cn

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  14. Show that C1^2 + 2C2^2 + 3C3^2 + ... + "^nCn^2 = ((2n-1!))/{(n-1)!}^2

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  15. Show that C2 + 2C3 + 3C4 + ... + (n-1)Cn = 1+ (n-2) 2^(n-1)

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  16. prove that :-C1 - 1/2 C2 + 1/3 C3 + ... + (-1)^(n+1) 1/n Cn = 1+1/2 + ...

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  17. C0 C1 + C1 C2 + .... + C(n-1) Cn = (2^n.n.1.3.5... (2n-1))/(n+1)

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  18. The sum 1/(1!9!) + 1/(3!7!) + ... + 1/(7!3!) + 1/(9!1!) can be written...

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  19. Using binomial theorem show that 1^(99) + 2^(99) +3^(99) + 4^(99) + 5^...

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  20. Using the binomial theorem show that 1^(99) + 2^(99) +3^(99) + 4^(99) ...

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