find the greatest term in the expansion of (10 +3x) ^12 when X =4
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The 12th term in the expansion of (10+3x)^12 will be numerically the greatest.
The greatest term in the expansion of (10+3x)^12 is calculated by using the properties of ratios. For the term T to be maximum, T(r+1) > T(r).
T(r+1) / T(r) = ((n – r + 1)/r) x (10x/3)
Substituting n = 12 and x = 4,
T(r+1) / T(r) = ((13 – r)/r) x (40/3)
For finding the greatest term, T(r+1) > T(r).
520 – 40r > 3r
520 > 43r
r < 520/43
r < 12.099
Taking r to be an integer, we calculate r to be 12.
Hence, the 12th term in the expansion of (10+3x)^12 will be numerically the greatest.
HOPE , IT HELPS ... ✌️
________________________
______________________
The 12th term in the expansion of (10+3x)^12 will be numerically the greatest.
The greatest term in the expansion of (10+3x)^12 is calculated by using the properties of ratios. For the term T to be maximum, T(r+1) > T(r).
T(r+1) / T(r) = ((n – r + 1)/r) x (10x/3)
Substituting n = 12 and x = 4,
T(r+1) / T(r) = ((13 – r)/r) x (40/3)
For finding the greatest term, T(r+1) > T(r).
520 – 40r > 3r
520 > 43r
r < 520/43
r < 12.099
Taking r to be an integer, we calculate r to be 12.
Hence, the 12th term in the expansion of (10+3x)^12 will be numerically the greatest.
HOPE , IT HELPS ... ✌️
sheoranprateek:
why here 10x/3 come not 10/3x
can u tell me orally
at last I understand that answer is 12th term but why 12c7(10)^5(12)^7come
pls reply quickly
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