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MathematicsSingle correct+4 / -1Question 9 of 75

JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 9

Let the smallest value of k∈Nk\in\mathbb N, for which the coefficient of x3x^3 in (1+x)3+(1+x)4+(1+x)5+⋯+(1+x)99+(1+kx)100(1+x)^3+(1+x)^4+(1+x)^5+\dots+(1+x)^{99}+(1+kx)^{100}, x≠0x\ne0, is (43n+1014)(100C3)\left(43n+\dfrac{101}4\right)\left({}^{100}C_3\right) for some n∈Nn\in\mathbb N, be pp. Then the value of p+np+n is:

Choose one option
Answer and worked solution

Official answer: B

Coefficient of x3x^3. ∑r=399(r3)+k3(1003)=(1004)+k3(1003)\displaystyle\sum_{r=3}^{99}\binom r3+k^3\binom{100}3=\binom{100}4+k^3\binom{100}3 (hockey-stick identity).

Simplify. (1004)=974(1003)\binom{100}4=\frac{97}4\binom{100}3, so the condition is 974+k3=43n+1014\frac{97}4+k^3=43n+\frac{101}4, i.e. k3−1=43nk^3-1=43n.

Smallest k. k=1k=1 gives n=0∉Nn=0\notin\mathbb N. The next solution of k3≡1(mod43)k^3\equiv1\pmod{43} is k=6k=6: 216−1=215=43⋅5216-1=215=43\cdot5.

Answer. p+n=6+5=11p+n=6+5=11.

(B)

Topics:Properties of binomial coefficientsNumber systems
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