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7.Binomial Theorem
hard
જો ${(1 + x)^{15}} = {C_0} + {C_1}x + {C_2}{x^2} + ...... + {C_{15}}{x^{15}},$ તો ${C_2} + 2{C_3} + 3{C_4} + .... + 14{C_{15}} = $
A
${14.2^{14}}$
B
${13.2^{14}} + 1$
C
${13.2^{14}} - 1$
D
એકપણ નહિ.
(IIT-1966)
Solution
(b) We have ${(1 + x)^{15}} = {C_0} + {C_1}x + {C_2}{x^2}. + …. + {C_{15}}{x^{15}}$
==> $\frac{{{{(1 + x)}^{15}} – 1}}{x} = {C_1} + {C_2}x + …. + {C_{15}}{x^{14}}$
Differentiating both sides with respect to $ x$,
we get $ = \frac{{x.15{{(1 + x)}^{14}} – {{(1 + x)}^{15}} + 1}}{{{x^2}}}$
= ${C_2} + 2{C_3}x + …… + {\,^{14}}{C_{15}}{x^{13}}$
Putting $x = 1$,
we get ${C_2} + 2{C_3} + …. + 14{C_{15}} = {15.2^{14}} – {2^{15}} + 1 = {13.2^{14}} + 1.$
Standard 11
Mathematics
