Proof of Sum of n terms in a G.P ?
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gp?
oh.. Geometric Progression
kk
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S = a + a r + a r^2 + a r^3 + ... + a r^n-1
= a (r^n - 1) /(r -1) r <> 1
Proof by Math Induction.
For n = 1, S = a = a (r^1 - 1)/(r - 1) . It is true
Let it be true for n = k.
Sk = a + ar+ ar^2+ a r^3...+ a r^(k-1) = a (r^k - 1 )/(r - 1)
S_k+1 = S_k + a r^k = a (r^k - 1) / (r - 1) + a r^k
= a [r^(k+1) - 1 ] / (r - 1)
Thus the predicate is true for k+1 also.
Hence proved.
= a (r^n - 1) /(r -1) r <> 1
Proof by Math Induction.
For n = 1, S = a = a (r^1 - 1)/(r - 1) . It is true
Let it be true for n = k.
Sk = a + ar+ ar^2+ a r^3...+ a r^(k-1) = a (r^k - 1 )/(r - 1)
S_k+1 = S_k + a r^k = a (r^k - 1) / (r - 1) + a r^k
= a [r^(k+1) - 1 ] / (r - 1)
Thus the predicate is true for k+1 also.
Hence proved.
click on red heart thanks above pls
how that formula came ????
you are just proving that it is applicable for any value of n
this is proof by induction. in that a formula is proposed. then it is shown that it is true... the formula is not derived... qn did not ask for derivation. it asked for proof...
you can derive like this:... S = a + r [ a + a r + a r^2 + a r^(n-2) ] ..... So S = a + r [ S - a r^(n-1) ] ..... S (r - 1) = a [r^n - 1] ...... So ... S = a [ r^n - 1 ] / [r - 1] .... thus proved...
Answered by
7
Given
S = a + r [ a + a r + a r^2 + a r^(n-2) ]
In the brackets we have n-1 terms of S itself.
Hence, S = a + r [ S - a r^(n-1) ]
=> S - r S = a - a r * r^(n-1)
Rearranging terms:
S (r - 1) = a (r^n - 1)
Finally, S = a ( r^n - 1) / (r - 1) or a (1 - r^n) / (1 - r)
S = a + r [ a + a r + a r^2 + a r^(n-2) ]
In the brackets we have n-1 terms of S itself.
Hence, S = a + r [ S - a r^(n-1) ]
=> S - r S = a - a r * r^(n-1)
Rearranging terms:
S (r - 1) = a (r^n - 1)
Finally, S = a ( r^n - 1) / (r - 1) or a (1 - r^n) / (1 - r)
GOOD :)
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